# SELECTED QUESTIONS

Some of the questions below are not the regular type of questions you will see in your assignments and tests. Don't worry if you haven't come across questions of this nature before; there's always a first time for everything. They are designed to make you get used to questions that require proofs, which is a useful skill to acquire, especially if you are going to do math beyond high school. We hope to add more questions, as the opportunity arises.

### On quadratics and their discriminants (Grades 10-12)

For a quadratic in the usual standard form $ax^2+bx+c$, the discriminant is the familiar quantity $$b^2-4ac$$ that appears in the quadratic formula. It is used to predict the nature of the roots without actually computing the roots.

1. If $b^2-2ac< 0$, PROVE that $b^2-4ac< 0$ also.
2. Observe that in order to have $b^2-2ac< 0$, $a$ and $c$ must have the same signs to begin with.

This exercise can be used as an alternative test for complex roots in a quadratic equation, deploying $b^2-2ac$ instead of the traditional $b^2-4ac$.

However, being (that $b^2-2ac< 0\implies b^2-4ac< 0$ is) a one-sided implication, this approach has some limitations. For example, it can happen that $b^2-2ac>0$ while $b^2-4ac< 0$ (check this for the quadratic $2x^2+4x+3$), so it is not entirely reliable.

As will be shown later, where the use of $b^2-2ac$ is more appropriate is when the degree is beyond $2$; that is, from cubics onwards. In those cases, the computation of the actual discriminant comes with increasing complexity, whereas $b^2-2ac$ is easy to calculate. So, if by chance $b^2-2ac< 0$, then we can infer that the polynomial contains complex roots, and the need to compute the actual discriminant may be obviated.

Finally, since the implications $$b^2-ac< 0~\implies b^2-4ac< 0 \quad\textrm{and}\quad b^2-3ac< 0~\implies b^2-4ac< 0$$ are also true, we will show in the next exercise why we have settled for $$b^2-2ac< 0~\implies b^2-4ac< 0,$$ instead of the other two.

3. If the roots of the quadratic equation $ax^2+bx+c=0$ are real, PROVE that $b^2-2ac> 0$.
4. What happens if $b^2-2ac=0$? We have mixed results, with all but one indicating complex roots. So, let's isolate this one case: it occurs when the two roots of the quadratic are zero, which only happens for the parent quadratic (equation $x^2=0$).

With that said, the contrapositive of the above statement is: if $b^2-2ac \leq 0$, then not all the roots are real. (Bear in mind what was said about $b^2-2ac=0$.)

Since a quadratic has only two roots and complex roots occur in conjugate pairs, $b^2-2ac< 0$ for a quadratic means that the quadratic has only complex roots.

This was why we chose $b^2-2ac< 0$ above.

5. For the standard form quadratic $ax^2+bx+c$, PROVE that:
• if the discriminant is an even perfect square, then $b$ must be an even number;
• if the discriminant is an odd perfect square, then $b$ must be an odd number.

The converse is not true in both cases.

One sees that the behaviour of the discriminant -- and the quadratic itself -- is, to some extent, influenced by $b$. Take $b$ for behaviour.

6. PROVE that if a quadratic $ax^2+bx+c$ has zero discriminant, then either $b=0$ and $c=0$, or $b\neq 0$ and $c\neq 0$. That is, either both $b$ and $c$ are zero simultaneously, or they are both non-zero.

In particular, a perfect square trinomial cannot have a missing constant term; otherwise, it will degenerate to the form $y=ax^2$, which doesn't directly fit into the trinomial definition.

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8. PROVE that $x=-\frac{b}{a}$ is a root of the quadratic equation $ax^2+bx+c=0$ if, and only if, $c=0$.

When put in the form $ax^2+bx+c$, the quadratics that go through the origin are precisely the ones with a root $-\frac{b}{a}$ (and of course $0$ as a second root).

9. Let $r,s$ be the roots of the quadratic equation $ax^2+bx+c=0$. PROVE that $b(rs)+c(r+s)=0$.
10. For a perfect square trinomial, this reduces to $br^2+2cr=0$ or $bs^2+2cs=0$, since $r=s$.

So we have $r(br+2c)=0$, whence $r=0$ or $r=\frac{-2c}{b}$.

Recall that the $x$-coordinate of the vertex of any quadratic $ax^2+bx+c$ is given by $x=\frac{-b}{2a}$. For a perfect square trinomial, the roots sit on the vertex, so $$\frac{-b}{2a}=\frac{-2c}{b}.$$ Re-arranging, we obtain $b^2=4ac$, the usual condition for a perfect square trinomial.

11. PROVE that if a quadratic expression can be given in factored/intercept form $y=a(x-r)(x-s), ~a\neq 0,$ then:
• its discriminant is $[a(r-s)]^2$
• its maximum/minimum value is $y=-a(\frac{r-s}{2})^{2}$

Actually, the first result applies to all quadratics, not only the ones in factored form.

So, here is the general statement, which we invite you to VERIFY: for any quadratic $ax^2+bx+c$ with roots $r,s$ (rational, irrational, or complex), the discriminant $\Delta$ can be given in terms of the roots as $\Delta=a^2(r-s)^2$.

Beware that this discriminant expression -- though appearing like a perfect square -- might in fact be negative, when the roots $r,s$ are complex numbers. From this it can be seen that the discriminant can always be dressed up like a perfect square, even when it is not. Don't be carried away by the external appearance of things; it is their internal constitution that matters most.

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13. Let $k\neq 0$ be a root of the quadratic equation $ax^2+bx+c=0$ (assume that $a,b,c\neq 0$ for convenience). PROVE that the discriminant $\Delta$ of the quadratic can then be given by $\Delta =(ak-\frac{c}{k})^{2}$.
14. Thus, the discriminant can also be expressed in terms of just one of the roots.

15. PROVE that for any quadratic (expect perfect square trinomials), the ratio of the discriminant to the maximum/minimum value is always equal to $-4a$.
16. PROVE that if a quadratic has a leading coefficient of $-\frac{1}{4}$ (i.e. $a=-\frac{1}{4}$ ), then the values of its discriminant and maximum are equal.

Except for perfect square trinomials, the above statement is biconditional.

Never underrate quadratics in which $a=-\frac{1}{4}$; they are quite special.

17. Let $a\neq 0$ and $n$ be integers. PROVE that the discriminant of the quadratic equation $$a^{n-1}x^2+(a^n+1)x+a=0$$ is always a perfect square, namely $(a^n-1)^2$.

Consequently, quadratics of the above form can be factored -- as $(x+a)(a^{n-1}x+1)$.

The case $a=1$ yields a perfect square trinomial, the only perfect square trinomial in this quadratic family, when $a$ is a real number.

An interesting case occurs when $a=2$, as the resulting discriminant is the square of a Mersenne number.

18. Let $a>1$ and $k>2$ be positive integers, with $k$ prime. PROVE that the discriminant of the quadratic equation $$a^{2k-1}x^2+(a^{2k}+1)x+a=0$$ is a product of four perfect squares.
19. The requirement that $a> 1$ is just to exclude the zero discriminant. Also, limiting $a$ to positive integer values is merely for convenience.

A proof that uses the principle of mathematical induction will be needed. The first value of $k$ to try, based on the conditions of the question, is $k=3$. The associated discriminant in this case is $(a^6-1)^{2}=(a-1)^2(a+1)^2(a^2+a+1)^2(a^2-a+1)^2$.

20. Let $a>1$ and $k$ be positive integers. PROVE that the discriminant of the quadratic equation $$a^{2k}x^2+(a^{2k+1}+1)x+a=0$$ is a product of two perfect squares.
21. Again, one uses the principle of mathematical induction in this exercise.

22. Let $a>1$ and $k$ be positive integers, with $k$ an even number of the form $k=2^{p}$. PROVE that the discriminant of the quadratic equation $$a^{2k-1}x^2+(a^{2k}+1)x+a=0$$ is a product of $(p+1)$ perfect squares.
23. Again, one uses the principle of mathematical induction in this exercise.

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24. Suppose that the quadratic $ax^2+bx+c$ has zero discriminant. PROVE that the associated quadratic $acx^2+(b^2-2ac)x+ac$ also has zero discriminant. The converse is also true.

Equivalently, the original quadratic being a perfect square trinomial implies that the new one is. This way, if the factorization of the first quadratic is $(px+q)^2$, then the second quadratic factors as $[pq(x+1)]^2$.

25. In furtherance of the question above, there is something more general in what follows. Let the discriminant of the quadratic $ax^2+bx+c~~(a,b,c\neq 0)$ be denoted by $\Delta_{1}$, and let the discriminant of the quadratic $acx^2+(b^2-2ac)x+ac$ be denoted by $\Delta_{2}$.
• PROVE that $\Delta_{2}=b^2\times\Delta_{1}$
• Hence, deduce that if one of the quadratics can be factored, then so is the other.
• Deduce also that if the discriminant of one of the quadratics is zero, then so is the discriminant of the other.
26. If the quadratic equation $ax^2+bx+c=0$ has roots $r$ and $s$, PROVE that the roots of the associated quadratic equation $acx^2+(b^2-2ac)x+ac=0$ are $-\frac{b}{c}r-1$ and $-\frac{b}{c}s-1$.

Again, one sees that if any of the two quadratics has equal roots, then so does the other. And if the former factors as $a(x-r)(x-s)$, then the latter factors as $ac(x+1+\frac{b}{c}r)(x+1+\frac{b}{c}s)$.

27. If the quadratic equation $ax^2+bx+c=0$ has roots $r$ and $s$ (with $r,s\neq 0$), PROVE that $ac(\frac{r}{s})^2+(b^2-2ac)(\frac{r}{s})+ac=0$ and that $ac(\frac{s}{r})^2+(b^2-2ac)(\frac{s}{r})+ac=0$.

Thus, any solution to $acx^2+(b^2-2ac)x+ac=0$ is a ratio of the solutions to $ax^2+bx+c=0$.

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### On quadratics and transformations (Grades 11 and 12)

1. PROVE that the discriminant of a quadratic function is invariant under a horizontal translation of the quadratic function. In other words, if $f(x)=ax^2+bx+c$ is the starting quadratic, PROVE that its discriminant is the same as that of $f(x+h)$, for $h> 0$ (or $h< 0$).
2. Since every $x$ coordinate moves by the same unit under a horizontal translation, the above exercise can be applied to the roots of a quadratic equation in particular. Thus, if the roots of two quadratic equations differ by the same amount, then their discriminants are equal.

To see this, assume that the quadratic equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$. VERIFY that the quadratic equation whose roots are $\alpha+k$ and $\beta +k$, for some constant $k$, is $$ax^2+(b-2ak)x+(c-bk+ak^2)=0,$$ and that the discriminant of this new equation is the same as the old one, both $b^2-4ac$.

The case when $k=\frac{b}{a}$ is worth a separate mention, as it leads to a different transformation.

3. Let $f(x)=ax^2+bx+c$ be a quadratic function. PROVE that another quadratic function $g(x)=Ax^2+Bx+C$ (with $A$ having same sign as $a$) is a reflection of $f(x)$ in the $y$-axis if, and only if, the roots of $g(x)$ differ from (i.e., exceed) those of $f(x)$ by $\frac{b}{a}$.
4. The above exercise may be taken as a characterization of horizontal reflection of quadratic functions.

The result assumes, implicitly, that the given quadratics have real roots. If the roots are complex numbers, it still holds.

The condition that $A$ has the same sign as $a$ is to remove the possibility of the second quadratic being a reflection of the first in both the $y$ and the $x$ axes.

5. PROVE that if two quadratics have the same discriminant and the same leading coefficient, then they are merely horizontal translations of each other. To do this, let $$f(x)=ax^2+bx+c,\quad \textrm{and}\quad g(x)=Ax^2+Bx+C$$ be the quadratics under consideration; for convenience, assume that $b,c,B,C\neq 0$. Set $h=\frac{B-b}{2a}=\frac{B-b}{2A}$, and PROVE that $f(x+h)=g(x)$, and hence $g(x-h)=f(x)$.
6. Beautiful, beautiful thing.

Notice that this is a partial converse to Question 1 above.

In general, if two quadratics have the same leading coefficient, then we need both a horizontal translation and a vertical translation to obtain one from the other. With equal discriminants, however, the need for a vertical translation is removed.

Thus, in terms of transformations, we can distinguish a special family of quadratics: the ones having the same discriminants and the same leading coefficients.

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7. PROVE that a horizontal stretch/compression and a vertical stretch/compression have the same effect on the discriminant of a quadratic function. In other words, if $f(x)=ax^2+bx+c$ is a quadratic function and $r\neq 0$ is a real number, then the discriminants of the transformed quadratics $f(rx)$ and $rf(x)$ are both equal to $r^2 \times(b^2-4ac)$.

If we take the discriminant as an operation $D$ on the set of quadratic functions, then $D$ enjoys $D(f(rx))=D(rf(x))=r^2\times D(f(x))$, where $f(x)=ax^2+bx+c$ and $r\neq 0$

8. PROVE that a quadratic moves, as its roots move. To this end, let $f(x)=ax^2+bx+c$ be a quadratic with roots $\alpha$ and $\beta$. If the roots are translated $h$ units, $h> 0$, to $\alpha +h$ and $\beta+h$, PROVE that the resulting quadratic $ax^2+(b-2ah)x+(c-bh+ah^2)$ is the same as the horizontal translation $f(x-h)$.
9. Translating just the roots of a quadratic by a certain amount is equivalent to a horizontal translation of the entire quadratic, by the same amount.

Well, this should be expected, because a quadratic is largely dependent on its roots.

10. Let $\alpha$ and $\beta$ be the roots of the quadratic equation $f(x)=ax^2+bx+c=0$. PROVE that the quadratic equation whose roots are $\frac{\alpha}{k}$ and $\frac{\beta}{k}$, for $k\neq 0$, is $k^2ax^2+kbx+c=0$, and that this is the same as the horizontal stretch or compression $f(kx)=0$.
11. Scaling just the roots of a quadratic by a certain factor is equivalent to a horizontal stretch or compression of the entire quadratic, by the same factor.

12. PROVE that the leading coefficient of a quadratic function is unaffected by linear operations on the roots of the quadratic. To this end, suppose that $\alpha$ and $\beta$ are the roots of a quadratic equation $ax^2+bx+c=0$. PROVE that the quadratic equation whose roots are $k\alpha$ and $k\beta$, for a non-zero integer $k$, is $ax^2+kbx+k^2c=0$, and the quadratic equation with roots $\alpha +k$ and $\beta+k$, is $ax^2+(b-2ak)x+(c-bk+ak^2)=0.$

Exponentiation of the roots does, however, have effect on the leading term.

Observe that $k$ doesn't have to be an integer in the above exercise; in that case, the conclusion has to be modified slightly, as follows: linear operations on the roots of a quadratic have a corresponding linear effect on the coefficients.

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### Number patterns and the discriminant (Grade 12)

Here we investigate how the discriminant of a quadratic ($ax^2+bx+c$) behaves if the coefficients $a,b,c$ form an arithmetic or a geometric progression. Then we investigate other number patterns.

This is purely for the interest of theory; you'll rarely see such numbers in actual numerical problems involving quadratics. What to note here is that as the coefficients grow, the discriminant tends to be negative.

1. Let $a,b,c$ be a geometric sequence. PROVE that the discriminant of the quadratic equation $ax^2+bx+c=0$ is negative.
2. Thus, if the coefficients in a quadratic equation form a geometric progression, the quadratic will not have real solutions.

For example, the quadratic equation $x^2+3x+9=0$ does not have real solutions because the coefficients $1,3,9$ are in a geometric progression; in fact, our one-sided discriminant $$b^2-2ac=3^2-2(1)(9)=-9< 0,$$ as well as the usual discriminant $b^2-4ac=3^2-4(1)(9)=-27< 0$.

Observe that the usual discriminant is a multiple of $3$ in this case. This is ALWAYS true if the integer coefficients in a quadratic form a geometric sequence.

3. Let $a,b,c$ be integers that are in a geometric progression. PROVE that the discriminant of the quadratic equation $ax^2+bx+c=0$ is a multiple of $3$.
4. We will encounter a similar result pertaining the discriminant of a cubic later.

By the way, quadratics in which the coefficients form a geometric progression have a special cubic connection.

5. Let $a,b,c$ be a geometric progression. PROVE that the quadratic $ax^2+bx+c$ can be converted into a DIFFERENCE OF CUBES or a SUM OF CUBES by multiplication with a binomial of the form $Ax+B$.
6. Simply find what the values of $A$ and $B$ should be, using the given information about $a,b,c$. Then carry out the expansion $(ax^2+bx+c)(Ax+B)$ to obtain a DIFFERENCE OF CUBES or a SUM OF CUBES.

Thus, without recourse to the cubic discriminant, we understand why cubic equations of the form $x^3-n^3=0$ and $x^3+n^3=0$ ALWAYS have ONLY ONE real solution: their factorizations contain a quadratic part in which the coefficients form a geometric progression.

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7. Let $a,b,c$ be an arithmetic sequence with a common difference $d$. If the discriminant of the quadratic equation $ax^2+bx+c=0$ is equal to zero, PROVE that $d=\pm\frac{\sqrt{3}}{2}b$.
[We want $a\neq 0$ as the only restriction.]
8. The case in which the coefficients form an arithmetic sequence is more acccommodating than when they form a geometric sequence, in which case the discriminant is non-negotiably negative.

Here's a little detail on how accommodating the arithmetic case can be. Suppose the coefficients $a,b,c$ are all positive as well as the common difference $d$, then the discriminant is positive when $d>\frac{\sqrt{3}}{2}b$, and negative when $d<\frac{\sqrt{3}}{2}b$.

For example, the coefficients in the quadratic $x^2+2x+3$ form an arithmetic sequence with a common difference of $1$; its discriminant -- $2^2-4(1)(3)=-8$ -- is negative because $d=1< \sqrt{3}=\frac{\sqrt{3}}{2}(2)$, that is $d< \frac{\sqrt{3}}{2}b$.

9. Let $ax^2+bx+c=0$ be a quadratic equation in which the (positive) integers $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that the discriminant of the quadratic is EQUAL TO ZERO if, and only if, $d=(3+2\sqrt{3})a$.
10. Alright. Next question please.

11. PROVE that it is IMPOSSIBLE to have a quadratic equation with ZERO DISCRIMINANT when the coefficients of the quadratic are INTEGERS that form an ARITHMETIC sequence.
12. Zero discriminant, integer coefficients, and arithmetic sequence cannot co-exist in the same quadratic environment.

13. Let $ax^2+bx+c=0$ be a quadratic equation in which the integer coefficients $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that the discriminant is a PERFECT SQUARE if $d=7a$.
14. Can we make the above statement biconditional? Not yet, in view of the next exercise.

15. Let $ax^2+bx+c=0$ be a quadratic equation in which the integer coefficients $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that the discriminant is a PERFECT SQUARE if $d=-a$.
16. This result is similar to the preceding one, but we have decided to separate them because the quadratics in question are two different families; in particular, when $d=-a$, the middle term is missing (because the coefficients become $a,0,-a$), giving rise to quadratic equations of the form $ax^2-a=0$.

Nevertheless, there is a transformation that connects these two quadratic families.

17. Let $ax^2+bx+c=0$ be a quadratic equation in which the integer coefficients $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that the quadratic that results from $d=7a$ and the one that results from $d=-a$ have the SAME DISCRIMINANT.
18. In the case when $d=7a$, the prototype is $f(x)=x^2+8x+15$, while $g(x)=x^2-1$ is the parent for the case when $d=-a$.

The two quadratics have the same leading coefficient and the same discriminant, so in view of a previous result (see the section on transformation of quadratics), there is a horizontal transformation that connects these two quadratics. In fact, let $$h=\frac{8-0}{2}=4,~\textrm{then}~g(x+h)=(x+4)^2-1=x^2+8x+15=f(x).$$ Similarly, $f(x-4)=(x-4)^2+8(x-4)+15=x^2-1=g(x)$.

Observe that we have used $h=\frac{b-B}{2a}$, where $b,B$ are the middle coefficients and $a$ is their common leading coefficient. This is the amount by which two quadratics with the same discriminant and the same leading coefficient get horizontally transformed to each other.

By the way, is $d=7a$ and $d=-a$ the only case when the discriminants of the resulting quadratics are equal? No way.

19. Let $a,b,c$ and $a,B,C$ be (two separate) arithmetic sequences with common differences $d_{1}$ and $d_{2}$, respectively. PROVE that the quadratics $ax^2+bx+c$ and $ax^2+Bx+C$ have the SAME DISCRIMINANTS if, and only if, $d_{1}+d_{2}=6a$.
20. Let's consider two numerical examples, while noting that of necessity we have $d_{1}\neq d_{2}$ above: the two sequences already have the same first terms, so they cannot afford to have the same common difference.

Take the quadratic $x^2+3x+5$, in which the coefficients $1,3,5$ form an arithmetic sequence with common difference $d_{1}=2$. Its discriminant is $3^2-4(1)(5)=9-20=-11$.

In order to construct another quadratic with the same discriminant, we need a $d_{2}$ that satisfies $d_{1}+d_{2}=6$; that is, $2+d_{2}=6$. We get $d_{2}=4$. The resulting quadratic is $x^2+5x+9$, with accompanying discriminant $5^2-4(1)(9)=25-36=-11$.

Second example. Let's take the quadratic $7x^2-x-9$, in which the coefficients $7,-1,-9$ form an arithmetic sequence with common difference $d_{1}=-8$. Its discriminant is $(-1)^2-4(7)(-9)=1+252=253$.

Another quadratic with the same discriminant (as $7x^2-x-9$) is obtained by first constructing a three-term arithmetic sequence with first term $7$ whose common difference $d_{2}$ satisfies $d_{1}+d_{2}=6(7)$; that is, $-8+d_{2}=42$. Since $d_{2}=50$, the resulting quadratic is $7x^2+57x+107$, with accompanying discriminant $57^2-4(7)(107)=3249-2996=253$.

Beautiful, useless thing.

21. Let $ax^2+bx+c=0$ be a quadratic equation in which the integer coefficients $a,b,c$ form an arithmetic sequence with common difference $d$. PROVE that its discriminant is a perfect square if, and only if, $d=-a$ or $d=7a$.
22. Perfect square discriminant, integer coefficients, and arithmetic sequence, are in short supply.

In fact, up to a horizontal translation, there is only one quadratic family in which the coefficients are integers that form an arithmetic sequence with the quadratic having a perfect square discriminant. It is the family we encountered earlier, namely $a(x^2+8x+15)$ (or its horizontal translation $a(x^2-1)$).

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For the next set of questions in this section, assume throughout that the coefficients $a,b,c$ of the quadratic $ax^2+bx+c$ are all positive AND that $a\geq 1$.

23. If $a,b,c$ are consecutive perfect squares, PROVE that the discriminant of the quadratic is negative.
24. The first few perfect squares are $1,4,9,16,\cdots$.

25. If $a,b,c$ are consecutive perfect cubes, PROVE that the discriminant of the quadratic is negative.
26. The first few perfect cubes are $1,8,27,64,\cdots$.

27. If $a,b,c$ are consecutive triangular numbers, PROVE that the discriminant of the quadratic is negative.
28. The fist few triangular numbers are $1,3,6,10,\cdots$. The $n$th triangular number is $\frac{n(n+1)}{2}$.

29. If $a,b,c$ are consecutive Fibonacci numbers, PROVE that the discriminant of the quadratic is negative.
30. Fibonacci numbers are generated via a recurrence relation: $$F_{n}=F_{n-1}+F_{n-2},\quad\quad F_{0}=0,~F_{1}=1.$$ So the first few Fibonacci numbers are $0,1,1,2,3,5,8,\cdots$. For our purposes, we don't want $0$ as the first Fibonacci number, in accordance with our restriction above where we stipulated that $a\geq 1$.

31. If $a,b,c$ are consecutive Lucas numbers, PROVE that the discriminant of the quadratic is negative.
32. The first few Lucas numbers are $2,1,3,4,7,\cdots$. In general, Lucas numbers are generated via a recurrence relation: $$L_{n}=L_{n-1}+L_{n-2},\quad\quad L_{0}=2,~L_{1}=1.$$

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### Game of discriminants (Grades 11 and 12)

When it comes to quadratics, one of the popular high school questions is to construct a quadratic equation with given roots.

In this section, we'll modify this tradition slightly, asking you to construct quadratic equations with given discriminants.

The questions here can be played as a GAME, and so they are suitable for a classroom activity. Two players (or two groups of students) will be needed. Player A will choose a random integer, and player B will produce a quadratic whose discriminant is the given integer (or will provide a proof that such a quadratic does not exist). Then player B will choose an integer, and player A will produce a quadratic whose discriminant is the chosen integer (or will give a proof that such a quadratic is not possible).

In order to give sufficient opportunities to the players, FIVE rounds will be needed. The winner is the player with more points. If there's a tie, the game can be repeated as the facilitator deems fit.

The players need to be versed in the underlying theory, which we present in the exercises below.

1. For any positive integer $k$, PROVE that there is a quadratic equation with discriminant $k$.
2. You may begin with the quadratic equation $x^2+3x+2=0$, whose discriminant is $1$.

Note that the coefficients of the resulting quadratic need not be integers, but the discriminant has to be.

3. Let $k$ be a PERFECT SQUARE. PROVE that there is a quadratic equation with INTEGER coefficients whose discriminant is $k$.
4. Note that this is related to the preceding question.

5. Let $k$ be a (positive) integer. PROVE that there is a quadratic equation with INTEGER coefficients whose discriminant is $k^2$.
6. Consider $x^2+(k+2)x+(k+1)=0$.

7. If the coefficients $a,b,c$ in the quadratic equation $ax^2+bx+c=0$ are all INTEGERS, PROVE that it is IMPOSSIBLE for the discriminant to equal $-1$.
8. Having $-1$ as discriminant strips the quadratic of integer coefficients.

Why does this happen?

Recall that the discriminant $\Delta$ is given by $\Delta=b^2-4ac$. If this is to equal $-1$, then $b$ MUST be an ODD number. Put $b=2n+1$ for some integer $n$. Then $$b^2-4ac=-1\implies (2n+1)^2-4ac=-1,\quad 4n^2+4n+2=4ac.$$ We can write $4n^2+4n+2$ as $2(2n^2+2n+1)$, where the expression in the bracket is ODD, so $4n^2+4n+2$ is at most a multiple of $2$. Now return to the equation $4n^2+4n+2=4ac$, and complete the argument that shows that BOTH $a$ and $c$ cannot be integers. Or, you can initiate an entirely different line of argument.

Is $-1$ the only discriminant that strips the quadratic of its INTEGER coefficients? NO.

9. If the coefficients $a,b,c$ in the quadratic equation $ax^2+bx+c=0$ are all INTEGERS, PROVE that it is IMPOSSIBLE for the discriminant to equal $2$.
10. What's going on here? We'll summarize it in the next two exercises.

11. If the coefficients $a,b,c$ in the quadratic equation $ax^2+bx+c=0$ are all INTEGERS, and the discriminant is an ODD number, PROVE that such a discriminant must be an ODD number of the form $4k+1$ (for some integer $k$).
12. Clearly, if the given ODD number is positive, we want $k$ to be a member of the set $\{0,1,2,3,\cdots\}$, and if it's a negative ODD number, then $k$ belongs to $\{\cdots,-3,-2,-1\}$.

We see why we couldn't get $-1$ as discriminant in a previous exercise: It's NOT possible to write $-1=4k+1$, with $k\in\{\cdots,-3,-2,-1\}$.

13. If the coefficients $a,b,c$ in the quadratic equation $ax^2+bx+c=0$ are all INTEGERS, and the discriminant is an EVEN number, PROVE that such a discriminant must be an EVEN number of the form $4k$ (for some integer $k$).
14. This explains why we couldn't get $2$ as discriminant in a previous exercise.

So, it is IMPOSSIBLE to get $\pm 2,\pm 6,\pm 10,\pm 14,\cdots$ as discriminants if ALL the coefficients in a quadratic equation are integers.

15. For any positive integer $k$, PROVE that there is a quadratic equation with INTEGER coefficients whose discriminant is $8k$.
16. Higher powers of $2$ beyond $8$ are also possible. So we can obtain quadratic equations (with INTEGER coefficients) whose discriminants are $16k, 32k, 64k,\cdots$, for any positive integer $k$.

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### Factoring with freedom and constraint (Grades 11 and 12)

A quadratic in the form $ax^2+bx+c$ can be factored over the rational numbers (which includes the integers) if its discriminant $b^2-4ac$ is a perfect square.

If all the coefficients $a,b,c$ are provided, then the factorability property can be decided -- by computing the discriminant.

In this section, any two of the coefficients $a,b,c$ will be FIXED (i.e., given) while the remaining one will be FREE. We investigate admissible values of the free coefficient that will guarantee factorability.

1. Consider the quadratic $ax^2+Bx+c$, in which the integer coefficients $a$ and $c$ are FIXED. PROVE that there are AT LEAST TWO values of $B$ for which the quadratic can be factored over the integers.
2. This is a problem on existence as well as enumeration. So we need to construct and count.

In the present case, one possible value of $B$ is $a+c$, giving the factorization $$ax^2+(a+c)x+c=(x+1)(ax+c).$$ What's the second permissible value of $B$?

If $a>1$ and $c> 1$, we can improve upon the above result.

3. Let $a>1$ and $c>1$ be integers. PROVE that there are AT LEAST FOUR values of $B$ for which the quadratic $ax^2+Bx+c$ can be factored over the integers.
4. Two such values are $B=a+c$ and $B=ac+1$, with accompanying factorizations $(x+1)(ax+c)$ and $(x+c)(ax+1)$, respectively.

5. Let $a$ and $c$ be prime numbers. PROVE that there are EXACTLY FOUR values of $B$ for which the quadratic $ax^2+Bx+c$ can be factored over the integers.
6. Recall that a prime number has only TWO DIVISORS: $1$ and itself.

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7. Let $a$ and $c$ be composite numbers such that $ac$ is NOT a perfect square. PROVE that the TOTAL number of admissible values of $B$ that will make the quadratic $ax^2+Bx+c$ factorable (over the integers) is THE SAME AS THE TOTAL NUMBER OF DIVISORS OF $ac$.
8. A number that is not prime is said to be composite.

9. Consider the quadratic $ax^2+Bx+c$, in which the integers $a$ and $c$ are fixed. PROVE that the TOTAL number of admissible values of $B$ that will make the quadratic factorable (over the integers) is ALWAYS AN EVEN NUMBER.
10. Take the quadratic $x^2+Bx+5$, for example. In order to be able to factor this over the integers, there are only two admissible values of $B$, namely $\pm 6$.

11. Consider the quadratic $ax^2+Bx+c$, in which the integers $a$ and $c$ are fixed. PROVE that the TOTAL number of admissible values of $B$ that will make the quadratic factorable (over the integers) is ONE MORE THAN THE TOTAL NUMBER OF DIVISORS OF $ac$ if, and only if, $ac$ is A PERFECT SQUARE.
12. Use the previous exercise and the fact that a perfect square has an odd number of divisors.

13. Of all the admissible values of $B$ that will make the quadratic $ax^2+Bx+c$ factorable ($a$ and $c$ are fixed), PROVE that the maximum such value is $ac+1$ if $a$ and $c$ have the same signs (and hence that the minimum of all those allowable values is $-ac-1$).
14. If $a$ and $c$ have opposite signs, then the maximum among all the allowable values of $B$ is $-ac-1$, and the minimum is $ac+1$.

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15. Consider the quadratic $ax^2+bx+C$, in which the integer coefficients $a$ and $b$ are FIXED. PROVE that there is an INFINITE choice for $C$ that guarantees factorability of the quadratic over the integers.
16. Let's start with a choice of $C$ that is likely to complete the square for the quadratic -- this is a popular high school quadratic question. It is $C=\frac{b^2}{4a}$. We then obtain the factorization $$ax^2+bx+\frac{b^2}{4a}=ax^2+\Big(\frac{b}{2}+\frac{b}{2}\Big)x+\frac{b^2}{4a}=\Big(ax+\frac{b}{2}\Big)\Big(x+\frac{b}{2a}\Big).$$ That there are an infinite possibilities for $C$ is because there are numerous ways to decompose the middle coefficient $b$ (we used $b=\frac{b}{2}+\frac{b}{2}$ above, but one can also use $\frac{b}{3}+\frac{2b}{3},~\frac{b}{4}+\frac{3b}{4},~\frac{b}{5}+\frac{4b}{5}, ~2b-b, 3b-2b$, and so on, obtaining a separate factorization for each decomposition). In general, the decomposition of $b$ can be accomplished via $b=kb+(1-k)b$, where $k$ is an integer or a rational number.

We can view the present problem as dealing with a product of sums, while a previous one was about a sum of products.

17. PROVE that quadratics of the form $ax^2+bx+b$ can be factored over the integers if, and only if, $b=4a$.
18. One thing to note about such quadratics is that they have a simple factorization character: they either factor as $a(x+2)^2$, or they do not factor at all -- over the integers.

19. PROVE that if the quadratic $ax^2+bx+c$ ($a,c\neq 0$) can be factored (over the rationals or reals), then so is the re-arrangement $cx^2+bx+a$. In fact, PROVE that if $ax^2+bx+c$ factors as $a(x-r)(x-s)$, with $r,s\neq 0$, then the factorization of $cx^2+bx+a$ will be $c(x-\frac{1}{r})(x-\frac{1}{s})$.

Thus, when the leading coefficient and the constant term of a quadratic equation are interchanged, the roots become inverted.

This also happens for a linear equation: if $a,b\neq 0$, then $ax+b=0$ and $bx+a=0$ have reciprocal solutions $\frac{-b}{a}$ and $\frac{-a}{b}$, respectively.

That's just about it, for degrees $1$ and $2$. In the case of a cubic or higher degree polynomial, more than one interchange is needed.

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### A reduction strategy (Grades 11 and 12)

Some strategic substitutions that simplify the process of solving quadratic and (some) higher degree polynomial equations will be explored in this section. The main idea is that some polynomial equations in which certain coefficients are missing may be easier to solve than the ones in which all coefficients are present.

In order to obtain the appropriate substitutions, one has to work backwards -- take a given polynomial function $f(x)$, apply a horizontal transformation $f(x-h)$ to it; then, in the transformed polynomial function, equate the unwanted coefficient to zero and solve for $h$ -- if possible. Now if $x$ in $f(x)$ is replaced by $x-h$, the unwanted term will be displaced.

1. Let $f(x)=ax^2+bx+c$ be a quadratic in which the middle term is non-zero, that is, $b\neq 0$. PROVE that there is a horizontal transformation that eliminates $b$, thus reducing the quadratic to the form $g(x)=ax^2+\frac{4ac-b^2}{4a}$.
2. Basically, what this reduction does is to move the vertex of the parabola from wherever it was to a point on the $y$-axis.

Observe that the reduced quadratic is easier to solve than the original one; however, this method of reducing a quadratic to a form where the middle term is missing is not a very efficient means of solving a quadratic equation, though it works.

3. PROVE that $ax^2+bx+c$ is a perfect square trinomial if, and only if, there is a substitution (horizontal transformation) that reduces it to the form $ax^2$.
4. Thus, among quadratics, only perfect square trinomials can be reduced to their simplest forms under the above substitution. The vertex becomes translated to the origin, in a single move.

5. PROVE that the quadratic equation $ax^2+bx+c=0$ admits real solutions if, and only if, the reduced form $ax^2+\frac{4ac-b^2}{4a}=0$ admits real solutions.
6. Both the original form and the reduced form have the same discriminant.

7. Let $ax^2+bx+c=0$ be a quadratic equation. PROVE that the two statements below are equivalent:
• the equation $ax^2+bx+c=0$ has real solutions
• there is a substitution that eliminates the constant term.
8. In a quadratic, the reduction strategy is always capable of eliminating the second term, but not the constant term.

In general, the farther, the harder. If the polynomial is of degree three or more, it becomes progressively difficult to eliminate terms beyond the second term. For example, to find an appropriate substitution that will eliminate the third term, one will have to solve a quadratic equation; to find a suitable substitution that will eliminate the fourth term, one will have to solve a cubic equation, and so on.

9. Let $ax^2+bx+c=0$ be a quadratic equation. PROVE that we can obtain the quadratic formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ without completing the square.
10. Beautiful, beautiful thing. The traditional method uses completing the square; the reduction method goes through a different route.

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11. Let $f(x)=ax^3+bx^2+cx+d$ be a cubic polynomial. PROVE that there is a horizontal transformation that eliminates the term containing $x^2$, reducing the cubic to the form $g(x)=Ax^3+Cx+D$, usually called the depressed cubic.
12. For a little bit of history regarding cubic equations, see this article.

This reduction strategy is usually called Cardano's method, and as the article stated, the original idea came from Scipione del Ferro and Tartaglia (independently of each other).

13. Let $f(x)=ax^3+bx^2+cx+d$ be a cubic function in which $b^2=3ac$. PROVE that there is a substitution that eliminates the $x^2$ and $x$ terms simultaneously, reducing the cubic to the form $g(x)=Ax^3+D$.
14. It follows that the general solution of such cubics are easier to write down.

15. Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ be a polynomial of degree $n$. PROVE that there is a substitution that eliminates the term containing $x^{n-1}$.
16. This is the general situation.

17. Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ be a polynomial of degree $n$, in which $\binom{n}{2}a_{n-1}^2=n^2(a_{n}a_{n-2})$. PROVE that there is a substitution that eliminates the $x^{n-1}$ and $x^{n-2}$ terms simultaneously.
18. This generalizes question 7 above.

19. PROVE that the polynomial $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ is a perfect $n$th power if, and only if, there is a substitution that reduces it to the form $a_{n}x^n$.
20. Both directions are immediate. If such a reduction is possible, then the polynomial has only one root -- namely $0$ -- in the reduced state. In view of the original substitution, the starting polynomial will have one repeated root. Conversely, if the polynomial is a perfect $n$th power, write it in the form $(Ax+C)^{n}$, then find a substitution that makes it become $ax^n$.

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### Perfect power polynomials (Grades 11 and 12)

As an application of the above reduction procedure, we derive some relationships among the coefficients of polynomials that have repeated factors -- repeated as many times as their degrees. Call them perfect power polynomials.

1. PROVE that the quadratic $ax^2+bx+c$ is a perfect square trinomial if, and only if, $b^2=4ac$.
2. Recall that the reduced form of a quadratic, under the usual substitution, is $ax^2 + \frac{4ac-b^2}{4a}$. Moreover, in a perfect power polynomial, only the leading term survives after reduction, so we must have $\frac{4ac-b^2}{4a}=0$, whence $b^2=4ac$. The other direction is also easy.

3. PROVE that $ax^3+bx^2+cx+d$ is a perfect cube quadrinomial if, and only if, $b^2=3ac$ and $bc=9ad$.
4. If $a$ is eliminated between $b^2=3ac$ and $bc=9ad$, the result is $c^2=3bd$. Thus, an equivalent requirement is that $b^2=3ac$ and $c^2=3bd$. Stated differently, the sequences $a,~b,~3c$ and $b,~c,~3d$ are both geometric.

5. PROVE that $ax^4+bx^3+cx^2+dx+e$ is a perfect fourth power quintinomial if, and only if, $b^3=16a^2d,~bd=16ae$, and $6b^2=16ac$.
6. The third condition, $6b^2=16ac$, can be reduced to $3b^2=8ac$. We have chosen not to reduce it to its simplest form for a purpose. There will be similar scenario in what follows.

7. PROVE that $ax^5+bx^4+cx^3+dx^2+ex+f$ is a perfect fifth power polynomial if, and only if, $be=25af,~10b^2=25ac,~ 2b^3=25a^2d$, and $b^4=125a^3e$.
8. As our manner is, we have chosen not to reduce $10b^2=25ac$. This is to allow the inquisitive reader to notice what is going on.

9. PROVE that $ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$ is a perfect sixth power polynomial if, and only if, $bf=36ag,~15b^2=36ac,~20b^3=216a^2d,~15b^4=1296a^3e$, and $b^5=1296a^4f$.
10. Observe that the number of conditions to be satisfied among the coefficients is one less than the degree of the polynomial. Observe also that these conditions are independent of each other.

11. PROVE that an $n$th degree polynomial $a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ is a perfect $n$th power if, and only if, the ($n-1$ independent) conditions $$a_{n-r}=\frac{\binom{n}{r}a_{n-1}^{r}}{n^{r}a_{n}^{r-1}},\quad\quad r=2,3,4,\cdots,n$$ are satisfied.
12. Beautiful, beautiful thing.

Each of these conditions -- expressible in terms of the leading coefficient $a_{n}$ and the coefficient of the second term $a_{n-1}$ -- is to be satisfied by a perfect power polynomial of degree $n$.

Observe that if $a_{n-1}=0$, then the above equation implies that all other subsequent coefficients are forced to be zero as well. Thus, in a perfect power polynomial of degree $n$, no term must be missing -- or else the polynomial reduces to the form $a_{n}x^{n}$. So we know immediately that something like $x^2+9$ is not a perfect square trinomial because the middle term is missing.

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### Sum of powers of roots of a quadratic (Grade 12)

For simplicity, we work with a quadratic equation in the usual standard form: $$ax^2+bx+c=0,\quad a\neq 0,\quad \textrm{with discriminant}\quad \Delta_{1} = b^2-4ac.$$ We also assume that the quadratic has roots $\alpha$ and $\beta$. We wish to construct new quadratic equations whose roots are positive integral powers of $\alpha$ and $\beta$, and see how the discriminants of the new quadratic equations relate to the original discriminant.

Unlike linear operations on the roots of a quadratic (which leave the leading term untouched), these exponential operations do affect the leading term, but more significantly, that middle man, $b$.

One more thing: the exercises in this section rely heavily on factoring and expanding, and you can be sure that they will stretch your factoring faculty. But do not worry.

1. PROVE that for a positive integer $n\geq 2$: $$\alpha^n+\beta^n=(\alpha+\beta)^n-\sum_{k=1}^{m}\binom{n}{k}\alpha^k\beta^k(\alpha^{n-2k}+\beta^{n-2k}),$$ where $m=\frac{n}{2}$ when $n$ is even and $m=\frac{n-1}{2}$ when $n$ is odd.
2. PROVE that:
• the quadratic equation with roots $\alpha^2$ and $\beta^2$ is given by $a^2x^2-(b^2-2ac)x+c^2=0$
• the discriminant of the new quadratic is $\Delta_{2}=b^2(b^2-4ac)$

Thus, the relationship between the discriminants in this case is $$\Delta_{2}=b^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.

3. PROVE that:
• the quadratic equation with roots $\alpha^3$ and $\beta^3$ is given by $a^3x^2+(b^3-3ac)x+c^3=0$
• the discriminant of the new quadratic is $\Delta_{3}=(b^2-ac)^2(b^2-4ac)$

Thus, the relationship between the discriminants in this case is $$\Delta_{3}=(b^2-ac)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.

4. PROVE that:
• the quadratic equation with roots $\alpha^4$ and $\beta^4$ is given by $a^4x^2-(b^4-4acb^2+2a^2c^2)x+c^4=0$
• the discriminant of the new quadratic is $\Delta_{4}=b^2(b^2-2ac)^2(b^2-4ac)$

Thus, the relationship between the discriminants in this case is $$\Delta_{4}=b^2(b^2-2ac)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.

5. PROVE that:
• the quadratic equation with roots $\alpha^5$ and $\beta^5$ is given by $a^5x^2+(b^5-5acb^3+5a^2c^2b)x+c^5=0$
• the discriminant of the new quadratic is $\Delta_{5}=(b^4-3acb^2+a^2c^2)^2(b^2-4ac)$

Thus, the relationship between the discriminants in this case is $$\Delta_{5}=(b^4-3acb^2+a^2c^2)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.

6. PROVE that:
• the quadratic equation with roots $\alpha^6$ and $\beta^6$ is given by $a^6x^2-(b^6-6acb^4+9a^2c^2b^2-2a^3c^3)x+c^6=0$
• the discriminant of the new quadratic is $\Delta_{6}=b^2(b^2-ac)^2(b^2-3ac)^2(b^2-4ac)$

Thus, the relationship between the discriminants in this case is $$\Delta_{6}=b^2(b^2-ac)^2(b^2-3ac)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.

7. PROVE that:
• the quadratic equation with roots $\alpha^7$ and $\beta^7$ is given by $a^7x^2+(b^7-7acb^5+14a^2c^2b^3-7a^3c^3b)x+c^7=0$
• the discriminant of the new quadratic is $\Delta_{7}=(b^6-5acb^4+6a^2c^2b^2-a^3c^3)^2(b^2-4ac)$

Thus, the relationship between the discriminants in this case is $$\Delta_{7}=(b^6-5acb^4+6a^2c^2b^2-a^3c^3)^2\times \Delta_{1}.$$ As expected, this is a perfect square multiple of the original discriminant.

8. It is evident that things can escalate quite quickly beyond exponent $7$, especially for that middle man, the coefficient of $x$. Despite this, is there a pattern in the successive coefficients of $x$, or in the progressive discriminants? Something is definitely happening.

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### Beyond quadratics: Cubics and co (Grade 12)

1. Let $\alpha,\beta,\gamma$ be real numbers, not all zero. PROVE that if $\alpha+\beta+\gamma=0$, then $\alpha\beta+\beta\gamma+\gamma\alpha\neq 0$.
2. In words, if the sum of three non-zero real numbers is zero, then the sum of the products of the numbers, taken two at a time, cannot be zero.

In the case of two numbers, this is very evident -- both numbers have to be non-zero in order to satisfy the specified condition, so the only possible product, namely $\alpha\beta$, is also non-zero.

Here's an example (not a proof!) in the case of three numbers: let $\alpha=2,\beta=0,\gamma=-2$. One has $$\alpha+\beta+\gamma=2+0+(-2)=0,$$ but $$\alpha\beta+\beta\gamma+\gamma\alpha= 2(0)+0(-2)+(-2)(2)=-4\neq 0.$$

What does the implication $\alpha+\beta+\gamma=0\implies\alpha\beta+\beta\gamma+\gamma\alpha\neq 0$ illustrate in relation to the roots of a cubic equation?

Well, it says that both the sum of roots and the sum of products of the roots, taken two at a time, cannot be zero simultaneously, provided ALL the roots are real.

What of a cubic equation like $x^3-1=0$, in which both ($\alpha+\beta+\gamma$ and $\alpha\beta+\beta\gamma+\gamma\alpha$) are zero, one might ask. What's happened here is that not all the roots are real; in fact, by explicit calculation, the roots are found to be $1,~\frac{-1\pm i\sqrt{3}}{2}$.

3. Let $\alpha,\beta,\gamma$ be real numbers, not all zero. PROVE that if $\alpha\beta+\beta\gamma+\gamma\alpha=0$, then $\alpha+\beta+\gamma\neq 0$.
4. Similar to the previous exercise. Aside: if $\alpha\beta+\beta\gamma+\gamma\alpha=0$, then there are just two possibilities: either ALL THREE ($\alpha,\beta,\gamma$) are non-zero or EXACTLY TWO of them are zero (remember that we don't want all three to be zero at the same time).

5. Let $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$ be real numbers, not all zero, such that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=0$. PROVE that the sum of the products of the numbers, taken two at a time, is non-zero; that is, $\sum_{i\neq j}\alpha_{i}\alpha_{j}\neq 0$.
6. Part of the underlying reason is that in any choice of the $\alpha_{i}$'s, at least two of them must be non-zero in order to satisfy the original hypothesis. Then, taking two of them at a time will eventually include the two non-zero member numbers at some point, which guarantees that at least one of the overall two-products will be non-zero, since the $\alpha_{i}$'s come from what is known as a field, or, more generally, an integral domain. Something else guarantees that the overall sum-of-two-products is non-zero.

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7. Let $\alpha, \beta,\gamma$ be the roots of the cubic equation $ax^3+bx^2+cx+d=0$. PROVE that if all the roots are real numbers, then $b^2-2ac\geq 0$.
8. What's the contrapositive of the above statement? If $b^2-2ac< 0$, then not all the roots of the cubic are real.

There you have it -- a condition for complex roots: $b^2-2ac< 0$. This works for any polynomial, and it saves us the stress of having to compute the discriminant -- however, there are drawbacks to this approach, so the usual discriminant is still dominant.

9. Suppose that all the roots $\alpha_{1},\alpha_{2},\alpha_{3},\cdots,\alpha_{n}$ of the $n$th degree polynomial equation $$a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}=0$$ are real numbers. PROVE that $$a_{n-1}^2-2a_{n}a_{n-2}\geq 0.$$
10. For a proof, use appropriate Vieta's formulas and a generalization of the identity $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$.

11. Let $a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}=0$ be any polynomial equation in which the FIRST THREE coefficients ($a_{n},a_{n-1},a_{n-2}$) form a geometric progression. PROVE that the polynomial contains complex roots.
12. There's an implicit assumption above, namely $a_{n},a_{n-1},a_{n-2}\neq 0$, that is, all the FIRST THREE terms are present.

Recall that we've already seen this result for quadratics, where the proof was accomplished by a direct computation of the quadratic discriminant.

In the more general case, our one-sided discriminant will be do, showing that the presence (or absence) of complex roots is hugely influenced by the FIRST THREE coefficients.

13. Let $a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}=0$ be any polynomial equation in which $a_{n-1}=0$ (that is, the second highest degree term is missing). If $a_{n}$ and $a_{n-2}$ have the same signs, PROVE that the polynomial contains complex roots.
14. Again we see how the FIRST THREE highest degree terms affect the nature of the roots of a polynomial equation.

A quadratic example: $x^2+9=0$. Compared with the standard form $ax^2+bx+c=0$, we have $a=1,b=0,c=9$. The second term is missing, and $a,c$ have the same signs. Our one-sided discriminant, in this case, is $b^2-2ac=0^2-2(1)(9)=-18 < 0$.

A cubic example: $x^3+2x-4=0$. This cubic contains a pair of complex roots due to the absence of the $x^2$ term and the fact that the $x^3$ term and the $x$ term have the same signs.

And so on.

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15. Let $a,d\neq 0$. If the cubic $ax^3+bx^2+cx+d$ is factorable (over the rational numbers), PROVE that so is the re-arrangement $dx^3+cx^2+bx+a$. In fact, if the factorization of $ax^3+bx^2+cx+d$ is $a(x-\alpha)(x-\beta)(x-\gamma)$, PROVE that $dx^3+cx^2+bx+a$ factors as $d(x-\frac{1}{\alpha})(x-\frac{1}{\beta})(x-\frac{1}{\gamma})$.
16. The assumption that $d\neq 0$ removes the possiblity of any of the roots $\alpha,\beta,\gamma$ being zero. Thus, we won't run into the problem of having to divide by zero in the reciprocals $\frac{1}{\alpha},~\frac{1}{\beta},~\frac{1}{\gamma}$.

17. If $\alpha,\beta,\gamma$ are the roots of the cubic equation $f(x)=ax^3+bx^2+cx+d=0$, PROVE that the cubic equation whose roots are $\alpha+h,~\beta+h,~\gamma+h$, for some real number $h$, is $$ax^3+(b-3ah)x^2+(c-2bh+3ah^2)x+(d-ch+bh^2-ah^3)=0,$$ and that this is the same as $f(x-h)=0$.
18. As in the case of a quadratic, translating just the roots of a cubic is equivalent to a horizontal translation of the entire cubic function. Similarly, scaling the roots in the form $\frac{\alpha}{k},~\frac{\beta}{k},~\frac{\gamma}{k}, ~k\neq 0$, is equivalent to the horizontal stretch or compression $f(kx)$.

19. PROVE that linear operations on the roots of a cubic equation do not alter the leading coefficient. In other words, if $\alpha,\beta,\gamma$ are the roots of the cubic $f(x)=ax^3+bx^2+cx+d=0$, then the cubic whose roots are $\alpha+h,~\beta+h,~\gamma+h$, and the cubic with roots $k\alpha,~k\beta,~k\gamma$ ($k$ an integer), all have the same leading coefficient as the original cubic, namely $a$.
20. If we remove the requirement that $k$ is an integer, then the conclusion will be restated: linear operations on the roots of a cubic have linear effect on the leading coefficients.

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21. PROVE that $x=-\frac{b}{a}$ is a solution of the cubic equation $ax^3+bx^2+cx+d=0$ if, and only, if $bc=ad$.
22. Let $b^2-4ac=0$. PROVE that $x=-\frac{b}{2a}$ is a solution of the cubic equation $ax^3+bx^2+cx+d=0$ if, and only if, $d=0$.
23. Suppose that $\alpha,~\beta,~\gamma,~\delta$ are the roots of the quartic equation $ax^4+bx^3+cx^2+dx+e=0$, PROVE that for some real number $h$, the quartic equation with roots $\alpha+h,~\beta+h,~\gamma+h,~\delta+h$ is given by $$ax^4+(b-4ah)x^3+(c-3bh+6ah^2)x^2+(d-2ch+3bh^2-4ah^3)x+(e-dh+ch^2-bh^3+ah^4)=0$$
24. Observe that when $h=0$, one reobtains the original quartic, as expected.

25. Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ be a polynomial of degree $n$. PROVE that $$f(x-h)=\sum_{k=0}^{n}\frac{f^{(k)}(-h)}{k!}x^k,$$ where $h>0$, for convenience.
26. Isn't this familiar from Taylor series? Almost.

One application of this formula is that we can use it to quickly write down the polynomial equation that results from translating the roots of the original polynomial $f(x)$ horizontally by $h$ units.

27. Let $f(x)=ax^4+bx^3+cx^2+dx+e$ be a quartic polynomial. PROVE that another quartic polynomial $g(x)=Ax^4+Bx^3+Cx^2+Dx+C$ is a reflection of $f(x)$ in the $y$-axis if, and only if, the roots of $g(x)$ differ from (i.e., exceed) those of $f(x)$ by $\frac{b}{2a}$.
28. It was assumed implicitly in the above exercise that the given quartics have real roots. The result still holds in the case of complex roots.

29. Let $f(x)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{1}x+a_{0}$ be an even degree polynomial. PROVE that a second polynomial of the same degree $g(x)=A_{n}x^n+A_{n-1}x^{n-1}+A_{n-2}x^{n-2}+\cdots+A_{1}x+A_{0}$ is a reflection of $f(x)$ in the $y$-axis if, and only if, the roots of $g(x)$ differ from (i.e., exceed) those of $f(x)$ by $\frac{2b}{na}$.
30. This may be taken as a characterization of horizontal reflection of even degree polynomials.

As in a previous exercise, we want the leading coefficients of both polynomials to have the same signs; this is to prevent the case where one of the polynomials is reflected in both the $y$ and $x$ axes, with the result still holding.

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### The discriminant of a cubic (Grade 12)

The notion of a discriminant extends beyond quadratics to cubics and higher degree polynomials, but the computations become progessively tedious as the degrees increase. For example, the discriminant of the cubic $ax^3+bx^2+cx+d$ is given by $$\Delta_{3}=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd;$$ it contains five terms, each homogeneous of degree $4$.

The exercises in this section are mostly a test of factoring and expanding ability. Assume throughout that the real numbers $a,b,c,d$ come from $ax^3+bx^2+cx+d$, and that $\Delta_{3}=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$.

1. If $a=b=c=d$, PROVE that $\Delta_{3}< 0$.
2. Such a cubic will have one pair of complex roots. In fact, it factors as $a(x+1)(x^2+1)$.

3. If the coefficients $a,b,c,d$ form a geometric sequence, PROVE that $\Delta_{3}< 0$.
4. As we saw before, this also happens in the quadratic case. Does it seem to hold in general?

5. If $a=-c$ and $b=-d$, PROVE that $\Delta_{3}=4(a^2-b^2)^2$.
6. The discriminant is a perfect square in this case; it can be made zero if we impose an extra condition.

7. If $a=c$ and $b=d$, PROVE that $\Delta_{3}=-4(a^2+b^2)^2$.
8. Compare this with the result of the previous question. The discriminant is always negative in this case, so the cubic contains a pair of complex roots -- in fact, it factors as $(ax+b)(x^2+1)$.

9. If $a=-b$ and $c=-d$, PROVE that $\Delta_{3}=4bc(b-c)^2$.
10. The discriminant is a multiple of a perfect square in this case. It can be made zero if $b=0$ or $c=0$ or $b=c$. However, $b=0$ is not allowed, in view of the fact that $a=-b$ from the onset. Why shouldn't $b$ be zero?

11. If $a=d$ and $b=c$, PROVE that $\Delta_{3}=(c+d)(c-3d)^{3}$.
12. By re-writing this discriminant as $(c+d)(c-3d)(c-3d)^2$, we see that it is again a multiple of a perfect square. It can be made zero with additional assumptions.

13. If $a=b$ and $c=d$, PROVE that $\Delta_{3}=-4ac(a+c)^2$.
14. In this case, the discriminant could be negative or positive, depending on whether $a$ and $c$ have the same or opposite signs.

15. If $b^2=3ac$ and $bc=9ad$, PROVE that $\Delta_{3}=0$.
16. This is expected, in view of the fact that $b^2=3ac$ and $bc=9ad$ are the conditions for the cubic to be a perfect cube quadrinomial (see the section on perfect power polynomials).

17. If $b^2-2ac< 0$, PROVE that $\Delta_{3}< 0$.
18. Recall that this was the case with quadratics as well. Does it seem to hold in general?

19. PROVE that the cubic $ax^3+bx^2+cx+d$ and its reduced/depressed form $$ax^3+(c-\frac{b^2}{3a})x+(d-\frac{bc}{3a}+\frac{2b^3}{27a^2})$$ have the same discriminant, namely $b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$.
20. Just compute the discriminant of the reduced cubic.

In passing, this result is expected, and is the reason why conclusions drawn from the roots of the reduced/depressed cubic also apply to the roots of the original cubic equation.

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21. If the coefficients $a,b,c,d$ of the cubic $ax^3+bx^2+cx+d$ are integers that form an arithmetic sequence, PROVE that the discriminant is a multiple of $8$.
22. Let's consider a few numerical examples as instances of the above result (be aware that this DOESN'T prove the result).

In the cubic $x^3+2x^2+3x+4$, the coefficients $1,2,3,4$ form an arithmetic sequence with a common difference of $1$. The discriminant is: $$2^2(3^2)-4(1)(3^3)-4(2^3)(4)-27(1^2)(4^2)+18(1)(2)(3)(4)=-200=-25\times 8.$$ The cubic $x^3-x-2$ has coefficients $1,0,-1,-2$ that form an arithmetic sequence with a common difference of $-1$. Its discriminant is: $$0^2(-1)^2-4(1)(-1)^3-4(0)^3(-2)-27(1)^2(-2)^2+18(1)(0)(-1)(-2)=-104=-13\times 8.$$ The cubic $2x^3-x^2-4x-7$ has coefficients $2,-1,-4,-7$ that form an arithmetic sequence with a common difference of $-3$. Its discriminant is: $$(-1)^2(-4)^2-4(2)(-4)^3-4(-1)^3(-7)-27(2)^2(-7)^2+18(2)(-1)(-4)(-7)=-5800=-725\times 8.$$ To prove the result, use the definition of the cubic discriminant and an enumeration of a general arithmetic sequence (e.g., $t,t+c,t+2c,t+3c$, where $c$ is the common difference) as the coefficients of the cubic.

23. If the coefficients $a,b,c,d$ of the cubic $ax^3+bx^2+cx+d$ are integers that form a geometric sequence, PROVE that the discriminant is a multiple of $16$.
24. To prove this, use the definition of the cubic discriminant and an enumeration of a general geometric sequence (e.g. $t,tr,tr^2,tr^3$, where $r$ is the common ratio) as the coefficients of the cubic.

25. PROVE that any cubic equation of the form $ax^3+d=0$ has exactly ONE real solution (and hence a pair of complex conjugate solutions).
26. Check that the cubic discriminant simplifies in this case, and is negative.

27. PROVE that the cubic equation $ax^3+bx^2+d=0$ has exactly ONE real solution (and hence a pair of complex conjugate solutions), provided $b$ and $d$ have the same signs.
28. Check that the cubic discriminant is negative under the condition that $b$ and $d$ have the same signs.

If $b$ and $d$ have opposite signs, it may still happen that the discriminant is negative, but this will depend on the actual magnitudes of the other coefficients.

29. PROVE that the (depressed) cubic equation $ax^3+cx+d=0$ has EXACTLY ONE real solution (and hence a pair of complex conjugate solutions), provided $a$ and $c$ have the same signs.
30. Check that the cubic discriminant is negative under the given conditions.

Alternatively, our one-sided discriminant $b^2-2ac$ may be used in this case; in fact, we have $b^2-2ac=0^2-2ac=-2ac< 0$, because $a$ and $c$ having the same signs guarantees that the product $ac$ is positive.

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### Dual degree polynomials (Grade 12)

At first sight, it seems not right, that a polynomial should possess two degrees. But we'll clarify this shortly.

Let's mention that the domain of the polynomials under consideration is the set of integers (for simplicity, we restrict to the positive integers), so that in a sense we're dealing with discrete functions that have internal polynomial character.

Suppose you are asked to find an equation that models the data presented in the table below:
n f(n)
1 1
2 4
3 9
you will most likely come up with the familiar quadratic function $f(n)=n^2$. What may be surprising is that the cubic function $$f(n)=\frac{1}{3}(n^3-3n^2+11n-6)$$ also works. In fact: $$f(1)=\frac{1}{3}\Big(1^3-3(1^2)+11(1)-6\Big)=\frac{1}{3}(1-3+11-6)=\frac{1}{3}(3)=1$$ $$f(2)=\frac{1}{3}\Big(2^3-3(2^2)+11(2)-6\Big)=\frac{1}{3}(8-12+22-6)=\frac{1}{3}(12)=4$$ $$f(3)=\frac{1}{3}\Big(3^3-3(3^2)+11(3)-6\Big)=\frac{1}{3}(27-27+33-6)=\frac{1}{3}(27)=9$$ So the function $f(n)$ possesses two degrees? Sounds weird? We say that $f$ has $2$ as its inherent or permanent degree, while $3$ is its instantaneous degree. It is IMPORTANT to note that the instantaneous degree works as long as we're dealing with JUST the data presented above; going a step further will alter it. For example, if we add one more row to the table above:
n f(n)
1 1
2 4
3 9
4 16
then the inherent degree of $f(n)$ remains as $2$, while its instantaneous degree changes to $4$.

Our main OBJECTIVE in this section is to make you practise with FRACTIONS, FACTORIALS, and FINITE DIFFERENCES.

1. PROVE that any two numbers $r$ and $s$ (where $r\neq s$) can be described by a LINEAR function, and by a QUADRATIC function. See the table below:
2. n f(n)
1 r
2 s

For the linear case, one obtains $$f(n)=(s-r)n+(2r-s);$$ for the quadratic case, $$f(n)=\frac{1}{2}\Big((s-r)n^2+(r-s)n+2r\Big).$$ ALWAYS ALWAYS ALWAYS remember that the instantaneous or temporary degree is limited to the data under consideration.

Let's also use this opportunity to mention that if the instantaneous degree is $2$, then the $n$-steps (or $x$-steps if you're more accustomed to $x$ and $y$) doesn't matter. In fact, if we change the $n$-step in the table above to two:
n f(n)
1 r
3 s
then the resulting quadratic model is $$f(n)=\frac{1}{2}\Big((s-r)n^2+(3r-3s)n+2s\Big).$$ How do these two quadratics relate? First, they have the same leading coefficients. But they also have the same discriminants: the discriminant of $\frac{1}{2}\Big((s-r)n^2+(r-s)n+2r\Big)$ is $$\frac{1}{4}\Big((r-s)^2-4(s-r)(2r)\Big)=\frac{1}{4}(r^2-2rs+s^2-8rs+8r^2)=\frac{1}{4}(9r^2-10rs+s^2),$$ and the discriminant of $\frac{1}{2}\Big((s-r)n^2+(3r-3s)n+2s\Big)$ is $$\frac{1}{4}\Big((3r-3s)^2-4(s-r)(2s)\Big)=\frac{1}{4}(9r^2-18rs+9s^2+8rs-8s^2)=\frac{1}{4}(9r^2-10rs+s^2).$$ So, by a previous exercise (see the section on transformation of quadratics), the two quadratics are merely horizontal translations of each other. Therefore, the $n$-steps (for the quadratic case) doesn't matter.

3. PROVE that any three numbers $r,s,t$ (with $2s\neq r+t$) can be described by a QUADRATIC function, and by a CUBIC function. See the table below:
4. n f(n)
1 r
2 s
3 t

You should obtain $$f(n)=\frac{1}{2}\Big((t+r-2s)n^2+(8s-5r-3t)n+(6r-6s+2t)\Big)$$ for the quadratic representation, and $$f(n)=\frac{1}{6}\Big((t+r-2s)n^3+(6s-3t-3r)n^2+(2s+2t-4r)n+(12r-6s)\Big)$$ for the cubic model.

We want you to use the idea of FINITE DIFFERENCES to derive these expressions, rather than substituting directly into the given expressions. This way, the purpose of this exercise will be realized.

At this point we should acknowledge that the polynomial having the instantaneous degree is often much more complicated in appearance -- and necessarily so -- than the actual, inherent polynomial. But that's the whole point of this exercise: we're mortgaging computational simplicity for our own conceptual interest and conceived objective. We want you to hone your skills in working with FRACTIONS, FACTORIALS, FINITE DIFFERENCES, and LINEAR SYSTEMS.

In the next set of exercises, we'll use instantaneous polynomials to describe some familiar number patterns.

5. PROVE that the first four prime numbers -- $2,3,5,7$ -- can be represented by a CUBIC polynomial and a QUARTIC polynomial. Use the table below:
6. n f(n)
1 2
2 3
3 5
4 7

The CUBIC model is $$f(n)=-\frac{1}{6}\Big(n^3-9n^2+14n-18\Big);$$ and the QUARTIC model is $$f(n)=-\frac{1}{24}\Big(n^4-6n^3-n^2+6n-48\Big).$$ As always, our goal is that you derive these expressions using FINITE DIFFERENCES, rather than verifying them by substitution.

Something somewhat interesting is that we can extend this procedure to obtain a polynomial formula for any number of consecutive prime numbers, but the HUGE (as in H-U-G-E) downside is that the computations are not practically implementable. Who is willing to derive a polynomial of degree $10$, for example?

7. PROVE that the first three perfect numbers -- $6,28,496$ -- can be described by a QUADRATIC equation. Use the table below:
8. n f(n)
1 6
2 28
3 496

You should obtain $f(n)=223n^2-647n+430$.

Note that a CUBIC model is also possible for the above data.

NOTICE that the coefficients $223,-647,430$ add up to $6$, the first number. This is a good way of checking if our model is correct.

9. PROVE that the first four perfect numbers -- $6,28,496,8128$ -- can be described by a CUBIC polynomial (as well as a QUARTIC polynomial). Use the table below:
10. n f(n)
1 6
2 28
3 496
4 8128

In this case we have $$f(n)=\frac{1}{6}\Big(6718n^3-38970n^2+70016n-37728\Big)$$ for the cubic model. We leave you to derive the QUARTIC counterpart.

NOTICE that the coefficients add up to $6$, as expected: $$\frac{1}{6}\Big(6718-38970+70016-37728\Big)=\frac{1}{6}(36)=6.$$ Useful for checking the correctness of our model.

11. PROVE that the first four terms -- $0,1,1,2$ -- of the Fibonacci sequence can be described by the cubic polynomial $$f(n)=\frac{1}{6}(2n^3-15n^2+37n-24).$$ Use the table below:
12. n f(n)
1 0
2 1
3 1
4 2

The Fibonacci example doesn't bode well for us, but it is good to be considered in view of our primary objective for this section.

In a nutshell it reveals the unavoidable complications that can arise with our approach. The entire Fibonacci sequence already has a simple formula that describes it, whereas our polynomial model of the first four terms is already cubic. NOT so nice.

13. PROVE that the first four terms of the geometric sequence $2^{n-1}$ (namely $1,2,4,8$) can be described by the cubic polynomial $$f(n)=\frac{1}{6}n(n^2-3n+8).$$ Use the table below:
14. n f(n)
1 1
2 2
3 4
4 8

What should we say?

15. PROVE that the first four terms of the geometric sequence $r^{n-1}$ (namely $1,r,r^2,r^3$, where $r$ is a positive integer) can be described by a cubic polynomial of the form $$f(n)=an^3+bn^2+cn+d.$$ Use the table below:
16. n f(n)
1 1
2 $r$
3 $r^2$
4 $r^3$

We have: $$a = \frac{(r-1)^3}{6}, \quad b=\frac{3(r-1)^2-6(r-1)^3}{6}, \quad \frac{-6+6r-9(r-1)^2+11(r-1)^3}{6}, \quad \frac{12-6r+6(r-1)^2-6(r-1)^3}{6}.$$ It follows that the constant term $d$ is ALWAYS an integer, namely $$2-r+(r-1)^2-(r-1)^3=(r-1)^0-(r-1)^1+(r-1)^2-(r-1)^3.$$ Incidentally, you'll now have an idea of the general formula for calculating $d$, up to any degree.

17. Let $r,s$ be a two-term sequence (where $r\neq s$). PROVE that:
• if $f(n)=an+b$ is its linear model, then $a+b=r$;
• if $f(n)=an^2+bn+c$ is its quadratic model, then $a+b+c=r$.
18. Useful for checking if our polynomial model is right.

19. Let $r,s,t$ be a three-term sequence (where $2s\neq r+t$). PROVE that:
• if $f(n)=an^2+bn+c$ is its quadratic model, then $a+b+c=r$;
• if $f(n)=an^3+bn^2+cn+d$ is its cubic model, then $a+b+c+d=r$.
20. Useful for checking if our polynomial model is right.

21. Let $b> 1$. PROVE that the $n$th (or last) finite difference for the exponential function $f(n)=b^n$ is $b(b-1)^{n}$.
22. In particular, the last finite difference for the exponential function $f(n)=2^n$ is $2$.

23. Let $b>1$. PROVE that the last finite difference for the exponential function $f(n)=b^n$ coincides with its constant ratio if, and only if, $b=2$.
24. For an exponential function of the form $f(n)=b^n$, the constant ratio between consecutive $f(n)$ values is always equal to $b$. Use this fact, and the preceding exercise.

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### On analytic geometry (Grades 10-12)

For more exercises on elementary analytic geometry, see our blog posts.

1. The vertices of a $\triangle ABC$ are located at the points $A(-2,-2),~B(2,0),~C(1,1)$ in the Cartesian plane. An altitude is drawn from vertex $C$ to meet $AB$ at point $H$. The right bisector of $AB$, drawn from point $M$ on $AB$, meets $AC$ at point $R$. PROVE that the lengths (of) $MR$ and $CH$ are reciprocals of each other. Specifically, PROVE that $$MR= \frac{\sqrt{5}}{3}=\frac{1}{\frac{3}{\sqrt{5}}} =\frac{1}{CH}$$
2. In the notation of Question 1 above -- save that we work with arbitrary coordinates this time -- PROVE that, for any triangle in which the altitude under consideration is internal, the ratio $\frac{CH}{MR}$ is always sandwiched between $1$ and $2$; specifically: $$1\leq \frac{CH}{MR}< 2$$

Equality in the left member of this inequality holds if and only if the triangle is isosceles. VERIFY this!

In passing, the point of the above exercise is that the ratio of the length of an altitude to the length of a right bisector -- suitably constructed -- is always greater than or equal to $1$, but strictly less than $2$.

3. PROVE that, in any triangle, a median always makes equal angles with an internal altitude and a right bisector.
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5. PROVE that the equation of the right bisector of the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ can be given by $$2(y_2-y_1)y+2(x_2-x_1)x=(x^2_{2}-x^{2}_{1}) +(y^2_{2}-y^{2}_{1})$$

There you have it -- beautiful, simple thing.

Especially for those who don't like going through the rigmarole of finding the equation of a right/perpendicular bisector of a line segment. You're welcome to use the above formula. Minor reductions may be necessary when using this formula; for example, one may have to divide by a common factor, or may have to re-write the equation in slope-intercept form, depending on the requirement of the question in question. You read that correctly, question in question.

6. Let $A(x_{1},y_{1}),~B(x_{2},y_{2}),~C(x_{3},y_{3})$ be the vertices of a $\triangle ABC$. PROVE that the equation of the altitude from vertex $A$ can be given by $$(y_{3}-y_{2})y+(x_{3}-x_{2})x=(y_{3}-y_{2})y_{1}+(x_{3}-x_{2})x_{1}$$

Quote this formula if you don't like the traditional procedure.

However, it is best to follow the step-by-step procedure, rather than to rely on closed formulas (which are not always possible, or if possible, may not be convenient).

NOTICE that one reobtains the right bisector equation via the substitutions $x_{1}=\frac{x_{2}+x_{3}}{2},~y_{1}=\frac{y_{2}+y_{3}}{2}$. What does this say about the relationship between an altitude and a right bisector?

7. Consider two lines with slopes $m_{1}$ and $m_{2}$. If the acute angle between the lines is $45^{\circ}$, PROVE that $$\textrm{either}\quad m_{2}=\frac{1+m_{1}}{1-m_{1}},\quad\textrm{or}\quad m_{1}=\frac{1+m_{2}}{1-m_{2}}.$$ Hence, deduce that a line with a slope of $1$ cannot make a $45^{\circ}$ angle with another line, unless the other line is parallel to the $y$-axis or the $x$-axis. Similarly for a line with $-1$ as slope.

This exercise gives us a way of constructing two lines that intersect at $45^{\circ}$ without using a protractor: Choose the slope of the first line in such a way that it is not equal to $1$, then use the equation above to determine the slope of the second line.

HINT: The (acute) angle $\theta$ between two lines with slopes $m_{1}$ and $m_{2}$ satisfies $\tan\theta=|\frac{m_{2}-m_{1}}{1+m_{2}m_{1}}|$, where $||$ denotes the usual absolute value.

8. Let $a,x,y$ be real numbers with $a\neq 0$. PROVE that any triangle with Cartesian coordinates $(x-a,~y),~(x+a,~y),~(x,~y+a\sqrt{3})$ is equilateral.

The presence of the irrational number $\sqrt{3}$ in the third coordinate says two things.

One, it supports a well-known fact: it is impossible to form an equilateral triangle in which all the coordinates are rational numbers. That's the price an equilateral triangle pays for having equal sides: sometimes, you won't have everything go your way, even when you try many angles like a triangle.

The second thing it says is that every irrational number that appears in the specification of the coordinates of an equilateral triangle must be bound to $\sqrt{3}$ somehow. For example, the points $(-\sqrt{5},0),~(\sqrt{5},0),~(0,\sqrt{15})$ form an equilateral triangle. On the surface, it appears that the coordinates are completely free of $\sqrt{3}$, but the fact that $\sqrt{15}=\sqrt{5}\times \sqrt{3}$ shows that $\sqrt{3}$ was embedded in one of the coordinates.

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### Counter-examples in high school math (Grade 12)

1. You will recall that logarithms do not always respect addition and subtraction, that is, $\log(x+y)\neq \log(x)+\log(y)$ and $\log(x-y)\neq \log(x)-\log(y)$, assuming all these logarithms are defined. In what follows, you are asked to prove that equality holds under certain conditions.
• PROVE that $\log(x+y)=\log(x)+\log(y)$ if, and only if, $y=\frac{x}{x-1}$, where $x>1$
• Requiring that $x\neq 1$ is enough for the rational expression, but not for the logarithm.

• PROVE that $\log(x-y)=\log(x)-\log(y)$ if, and only if, $x=\frac{y^2}{y-1}$, where $y>1$
• As above, restricting $y\neq 1$ is enough for the rational expression, but not for the logarithm.

2. Something similar to the above happens in the case of exponents: $a^{x+y}\neq a^x+a^y$ and $a^{x-y}\neq a^x-a^y$. However, equality holds in certain cases.
• PROVE that $a^{x+y}=a^x+a^y$ if, and only if, $x=\log_{a}(\frac{a^y}{a^y-1})$.
• The case when $a=2,~x=1,~y=1$ is a special case.

• PROVE that $a^{x-y}=a^x-a^y$ if, and only if, $x=\log_{a}(\frac{a^{2y}}{a^y-1})$.
• The case of $a=2,~x=2,~y=1$ is a special case.

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### On linear relations (Grade 9)

• PROVE that the line whose equation is given in standard/general form as $$ax+by+c=0,\quad c\neq 0$$ cannot contain a point with coordinates $(\frac{a}{c},\frac{b}{c})$.
• PROVE that the equation of a line which goes through the points $(r,s)$ and $(s,r)$ can be given by $$x+y=r+s$$
• PROVE that the slope of a line
• is positive if and only if the $x$ and $y$ intercepts have opposite signs
• is negative if and only if the $x$ and $y$ intercepts have the same signs
• is equal to $-1$ if and only if the $x$ and $y$ intercepts are equal

Aside: It has been suggested that freedom from boredom can be obtained by doing something different from the norm, something different from your curriculum. This is another reason behind these exercises: If you feel bored to the core, feel free to take these exercises as a chore; they may be able to bring you a cure -- for boredom.