Math Core

Many AMC quadratic problems never ask you to find the roots. They ask for something built from the roots, like the sum of their squares or a new equation whose roots are related. Vieta's formulas let you answer those questions using only the coefficients, often with no square roots in sight.

Sum and product of the roots

If $r$ and $s$ are the roots of $ax^2 + bx + c = 0$ , then the quadratic factors as $a(x - r)(x - s)$ . Expanding gives

a(x - r)(x - s) = ax^2 - a(r + s)x + a\,rs.

Match this with $ax^2 + bx + c$ term by term.

Vieta's formulas for a quadratic

If $r$ and $s$ are the roots of $ax^2 + bx + c = 0$ , then

r + s = -\dfrac{b}{a} \qquad \text{and} \qquad rs = \dfrac{c}{a}.

These hold whether the roots are integers, irrational or even complex.

The graph of $y = x^2 - 4x + 1$ crosses the $x$ -axis at $2 - \sqrt{3}$ and $2 + \sqrt{3}$ . Their sum is $4$ and their product is $4 - 3 = 1$ , exactly as Vieta predicts.

The roots of x² − 4x + 1 add to 4 and multiply to 1.Open in grapher →

Symmetric expressions

An expression in $r$ and $s$ is symmetric if swapping $r$ and $s$ leaves it unchanged, like $r^2 + s^2$ or $\dfrac{1}{r} + \dfrac{1}{s}$ . Every symmetric expression can be rewritten using only $r + s$ and $rs$ . These are the identities to know by heart:

Expression	In terms of $r + s$ and $rs$
$r^2 + s^2$	$(r + s)^2 - 2rs$
$\dfrac{1}{r} + \dfrac{1}{s}$	$\dfrac{r + s}{rs}$
$(r - s)^2$	$(r + s)^2 - 4rs$
$r^3 + s^3$	$(r + s)^3 - 3rs(r + s)$
$r^4 + s^4$	$(r^2 + s^2)^2 - 2(rs)^2$

Notice that $(r - s)^2 = \dfrac{b^2 - 4ac}{a^2}$ . That is where the discriminant comes from: the roots are equal exactly when $b^2 - 4ac = 0$ , and they are real exactly when $b^2 - 4ac \ge 0$ .

Worked example: Symmetric expressions

Let $r$ and $s$ be the roots of $x^2 - 7x + 3 = 0$ . Find $r^2 + s^2$ and $\dfrac{1}{r} + \dfrac{1}{s}$ .

By Vieta, $r + s = 7$ and $rs = 3$ . Then

r^2 + s^2 = 7^2 - 2 \cdot 3 = 43, \qquad \frac{1}{r} + \frac{1}{s} = \frac{r + s}{rs} = \frac{7}{3}.

The roots themselves are $\dfrac{7 \pm \sqrt{37}}{2}$ , which you never needed.

Building a new quadratic

To write a quadratic whose roots are $u$ and $v$ , you only need $u + v$ and $uv$ : the monic quadratic is $x^2 - (u + v)x + uv$ . So if the new roots are built from old roots, compute the new sum and product with Vieta.

Worked example: Transformed roots

The roots of $x^2 - 5x + 2 = 0$ are $r$ and $s$ . Find a quadratic with integer coefficients whose roots are $r + \dfrac{1}{s}$ and $s + \dfrac{1}{r}$ .

We have $r + s = 5$ and $rs = 2$ .

New sum: $\left(r + \dfrac{1}{s}\right) + \left(s + \dfrac{1}{r}\right) = (r + s) + \dfrac{r + s}{rs} = 5 + \dfrac{5}{2} = \dfrac{15}{2}$ .

New product: $\left(r + \dfrac{1}{s}\right)\left(s + \dfrac{1}{r}\right) = rs + 1 + 1 + \dfrac{1}{rs} = 2 + 2 + \dfrac{1}{2} = \dfrac{9}{2}$ .

So the quadratic is $x^2 - \dfrac{15}{2}x + \dfrac{9}{2}$ , or, multiplying by $2$ , $2x^2 - 15x + 9 = 0$ .

Integer roots

When a problem says the roots are integers, write the two Vieta equations and eliminate the parameter. What is left usually factors with a "add a constant to both sides" trick.

Worked example: A parameter with integer roots

For how many integers $k$ does $x^2 - kx + (k + 11) = 0$ have two integer roots (possibly equal)? What is the sum of all such $k$ ?

Let the roots be $r$ and $s$ . Vieta gives $r + s = k$ and $rs = k + 11$ . Subtract to eliminate $k$ :

rs - r - s = 11 \quad\Longrightarrow\quad rs - r - s + 1 = 12 \quad\Longrightarrow\quad (r - 1)(s - 1) = 12.

The unordered factor pairs of $12$ are $(1, 12), (2, 6), (3, 4)$ and their negatives $(-1, -12), (-2, -6), (-3, -4)$ . Since $k = r + s = (r - 1) + (s - 1) + 2$ , the values are

k = 15,\ 10,\ 9,\ -11,\ -6,\ -5.

There are $6$ values, and their sum is $12$ . (Check $k = 9$ : $x^2 - 9x + 20 = (x - 4)(x - 5)$ .)

Common mistake

Watch the signs. For $ax^2 + bx + c$ , the sum of the roots is $-\dfrac{b}{a}$ , not $\dfrac{b}{a}$ . And if the leading coefficient isn't $1$ , you must divide by it: the roots of $2x^2 - 6x - 5$ add to $3$ , not $6$ .

Tip

If a problem says "real roots," check the discriminant at the end. Vieta happily produces a sum and product for complex roots too, and a contest answer choice may be designed to catch that.

Practice

Practice 1

Let $r$ and $s$ be the roots of $2x^2 - 6x - 5 = 0$ . What is $r^2 + s^2$ ?

Practice 2

The equation $x^2 + bx + 18 = 0$ has two roots, one of which is twice the other. If $b > 0$ , what is $b$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Let $r$ and $s$ be the roots of $x^2 - 3x + 1 = 0$ . What is $r^4 + s^4$ ?

Practice 4

The two roots of $x^2 - kx + 12 = 0$ differ by $1$ , and $k > 0$ . What is $k$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

The roots of $x^2 + 2x - 4 = 0$ are $r$ and $s$ . Which quadratic has roots $r^2$ and $s^2$ ?

Practice 6

Both roots of $x^2 - 21x + c = 0$ are prime numbers. What is $c$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

For some integers $m$ , both roots of $x^2 + mx + (m + 7) = 0$ are integers. What is the sum of all such $m$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

For which value of $k$ do the equations $x^2 + kx + 1 = 0$ and $x^2 + x + k = 0$ have a common real root?

Real contest practice

2005 AMC 10A, Problem 10: a quadratic with exactly one solution, using the discriminant.
2021 AMC 12A, Problem 12: Vieta's formulas for a higher-degree polynomial with integer roots.
2015 AMC 10A, Problem 23: a quadratic with a parameter whose zeros must be integers.