Sum and Product of Roots

You can find the sum and the product of all the roots of a polynomial equation without solving it. They come straight from the coefficients. This is a big time-saver: it lets you find missing coefficients, find the other roots once you know one, and build new equations whose roots are related to the old ones, all without ever using the quadratic formula.

Key ideas

Where the rule comes from: quadratics

If a quadratic $ax^2 + bx + c$ has roots $\alpha$ and $\beta$, then by the factor theorem it factors as $a(x - \alpha)(x - \beta)$. Expand that:

$$a(x - \alpha)(x - \beta) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$$

Compare the coefficients with $ax^2 + bx + c$: $\quad b = -a(\alpha + \beta)$ and $c = a\alpha\beta$. So

$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

The general rule

For a polynomial equation of degree $n$,

$$a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0, \qquad a_n \ne 0,$$

the $n$ roots (counted with repeats, and including complex roots) have

$$\text{sum} = \frac{-a_{n-1}}{a_n}, \qquad \text{product} = \frac{(-1)^n a_0}{a_n}$$

So you only ever need three coefficients: the leading one $a_n$, the next one $a_{n-1}$, and the constant $a_0$.

Degree	Equation	Sum of roots	Product of roots
2	$ax^2 + bx + c = 0$	$-\dfrac{b}{a}$	$\dfrac{c}{a}$
3	$ax^3 + bx^2 + cx + d = 0$	$-\dfrac{b}{a}$	$-\dfrac{d}{a}$
4	$ax^4 + bx^3 + cx^2 + dx + e = 0$	$-\dfrac{b}{a}$	$\dfrac{e}{a}$

The sign of the product alternates with the degree: positive for even degree, negative for odd degree.

Watch for missing terms

If a power is missing, its coefficient is $0$. For $x^4 - 3x^3 + 7 = 0$, the constant is $a_0 = 7$, and for $2x^3 + 5x - 4 = 0$ the $x^2$ coefficient is $0$, so the sum of the roots is $0$.

Complex roots

The rule counts all $n$ roots, real and complex. If the coefficients are real, complex roots come in conjugate pairs $p + qi$ and $p - qi$. A pair has a real sum and product, which makes the rule very handy:

$$(p + qi) + (p - qi) = 2p, \qquad (p + qi)(p - qi) = p^2 + q^2$$

See complex conjugate roots for more.

Useful identities for quadratics

Many expressions in $\alpha$ and $\beta$ can be rewritten using only $\alpha + \beta$ and $\alpha\beta$:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta, \qquad \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$$

Forming an equation with related roots

A quadratic with roots $r_1$ and $r_2$ is

$$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$$

that is, $x^2 - (\text{sum})x + (\text{product}) = 0$. Multiply through to clear fractions if you want integer coefficients.

For roots shifted or scaled from the old ones, a substitution also works for any degree. If the new roots are $\alpha + 1$, then a new root $x$ satisfies $x = \alpha + 1$, so $\alpha = x - 1$: replace $x$ by $x - 1$ in the old equation. If the new roots are $2\alpha$, replace $x$ by $\dfrac{x}{2}$.

(The full set of relationships between roots and coefficients, called Vieta’s formulas, also includes sums of products of pairs of roots. The IB guide only requires the sum and the product.)

Worked examples

Example 1: A quadratic

The roots of $2x^2 - 7x + 4 = 0$ are $\alpha$ and $\beta$. Without solving the equation, find $\alpha + \beta$, $\alpha\beta$ and $\alpha^2 + \beta^2$.

Solution. Here $a = 2$, $b = -7$, $c = 4$:

$$\alpha + \beta = -\frac{-7}{2} = \frac{7}{2}, \qquad \alpha\beta = \frac{4}{2} = 2$$ $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{49}{4} - 4 = \frac{33}{4}$$

Example 2: A cubic and a quartic

Find the sum and the product of the roots of

• (a) $3x^3 - 6x^2 + x + 12 = 0$
• (b) $2x^4 - 5x^3 + x - 6 = 0$

Solution.

(a) Degree $n = 3$, with $a_3 = 3$, $a_2 = -6$, $a_0 = 12$:

$$\text{sum} = \frac{-(-6)}{3} = 2, \qquad \text{product} = \frac{(-1)^3(12)}{3} = -4$$

(b) Degree $n = 4$, with $a_4 = 2$, $a_3 = -5$, $a_0 = -6$. There’s no $x^2$ term, but that coefficient isn’t needed:

$$\text{sum} = \frac{-(-5)}{2} = \frac{5}{2}, \qquad \text{product} = \frac{(-1)^4(-6)}{2} = -3$$

These totals include any complex roots. (The cubic in (a) has one real root and two complex ones, but you didn’t need to know that.)

Example 3: Using a complex root

The equation $x^3 + px^2 + qx - 10 = 0$, where $p, q \in \mathbb{R}$, has a root $1 + 2i$. Find the other roots, and the values of $p$ and $q$.

Solution. The coefficients are real, so the conjugate $1 - 2i$ is also a root. Call the third root $\gamma$.

Product of roots (degree $3$, $a_0 = -10$):

$$(1 + 2i)(1 - 2i)\gamma = \frac{(-1)^3(-10)}{1} = 10 \quad\Rightarrow\quad 5\gamma = 10 \quad\Rightarrow\quad \gamma = 2$$

Sum of roots:

$$(1 + 2i) + (1 - 2i) + 2 = 4 = -\frac{p}{1} \quad\Rightarrow\quad p = -4$$

Find $q$ by multiplying out the factors. The conjugate pair gives $(x - 1)^2 + 4 = x^2 - 2x + 5$, so

$$(x - 2)(x^2 - 2x + 5) = x^3 - 4x^2 + 9x - 10$$

So the roots are $2$ and $1 \pm 2i$, with $p = -4$ and $q = 9$. Check: the expansion gives $p = -4$ ✓ and constant $-10$ ✓.

Example 4: An equation with roots one more

The roots of $2x^2 - 3x - 4 = 0$ are $\alpha$ and $\beta$. Find a quadratic equation with integer coefficients whose roots are $\alpha + 1$ and $\beta + 1$.

Solution. First, $\alpha + \beta = \dfrac{3}{2}$ and $\alpha\beta = \dfrac{-4}{2} = -2$.

Method 1: sum and product.

$$\text{new sum} = (\alpha + 1) + (\beta + 1) = \frac{3}{2} + 2 = \frac{7}{2}$$ $$\text{new product} = (\alpha + 1)(\beta + 1) = \alpha\beta + (\alpha + \beta) + 1 = -2 + \frac{3}{2} + 1 = \frac{1}{2}$$

So the equation is $x^2 - \dfrac{7}{2}x + \dfrac{1}{2} = 0$. Multiply by $2$:

$$2x^2 - 7x + 1 = 0$$

Method 2: substitution. Replace $x$ by $x - 1$ in the original equation:

$$2(x - 1)^2 - 3(x - 1) - 4 = 2x^2 - 4x + 2 - 3x + 3 - 4 = 2x^2 - 7x + 1$$

Both methods give $2x^2 - 7x + 1 = 0$. ✓

Common mistakes

Forgetting the minus sign in the sum. The sum is $-\dfrac{b}{a}$, not $\dfrac{b}{a}$. For $x^2 - 5x + 6 = 0$ the roots are $2$ and $3$, with sum $5$, which is $-\dfrac{-5}{1}$.

Getting the sign of the product wrong for odd degrees. The product has the factor $(-1)^n$. For a cubic, it’s $-\dfrac{d}{a}$. Check with an easy case: $(x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$ has root product $6 = -\dfrac{-6}{1}$.

Forgetting to divide by the leading coefficient. If $a_n \ne 1$, every formula has $a_n$ in the denominator. $2x^2 - 7x + 4 = 0$ has product $2$, not $4$.

Using the wrong coefficient when a term is missing. In $2x^4 - 5x^3 + x - 6 = 0$, the coefficient you need for the sum is that of $x^3$, which is $-5$, not the next non-zero one after it. If the $x^{n-1}$ term is missing, the sum is $0$.

Forgetting complex roots. The sum and product include every root, real or not. A cubic with only one real root still has three roots in the formula.

Squaring the sum instead of using the identity. $\alpha^2 + \beta^2$ is not $(\alpha + \beta)^2$. You must subtract $2\alpha\beta$.

Practice

1. (Warm-up) Find the sum and the product of the roots of $5x^2 + 3x - 2 = 0$.

Solution

$$\text{sum} = -\frac{3}{5}, \qquad \text{product} = \frac{-2}{5} = -\frac{2}{5}$$

Check by solving: $(5x - 2)(x + 1) = 0$ gives roots $\dfrac{2}{5}$ and $-1$, with sum $-\dfrac{3}{5}$ ✓ and product $-\dfrac{2}{5}$ ✓.

2. (Warm-up) Find the sum and the product of the roots of $2x^3 + x^2 - 8x + 6 = 0$.

Solution

Degree $3$, $a_3 = 2$, $a_2 = 1$, $a_0 = 6$:

$$\text{sum} = -\frac{1}{2}, \qquad \text{product} = \frac{(-1)^3(6)}{2} = -3$$

3. (Warm-up) Find the sum and the product of the roots of $x^4 - 3x^3 + 2x^2 + 7 = 0$.

Solution

Degree $4$, $a_4 = 1$, $a_3 = -3$, $a_0 = 7$:

$$\text{sum} = \frac{-(-3)}{1} = 3, \qquad \text{product} = \frac{(-1)^4(7)}{1} = 7$$

4. (Core) The roots of $x^2 - 4x + 1 = 0$ are $\alpha$ and $\beta$. Find the exact values of $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ and $\alpha^2 + \beta^2$.

Solution

$\alpha + \beta = 4$ and $\alpha\beta = 1$.

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{4}{1} = 4$$$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 16 - 2 = 14$$

5. (Core) The roots of $3x^2 - 5x + 1 = 0$ are $\alpha$ and $\beta$. Find a quadratic equation with integer coefficients whose roots are $2\alpha$ and $2\beta$.

Solution

$\alpha + \beta = \dfrac{5}{3}$ and $\alpha\beta = \dfrac{1}{3}$.

New sum: $2\alpha + 2\beta = \dfrac{10}{3}$. New product: $(2\alpha)(2\beta) = 4\alpha\beta = \dfrac{4}{3}$.

$$x^2 - \frac{10}{3}x + \frac{4}{3} = 0 \quad\Rightarrow\quad 3x^2 - 10x + 4 = 0$$

Check by substitution: replace $x$ by $\dfrac{x}{2}$ in the original: $\dfrac{3x^2}{4} - \dfrac{5x}{2} + 1 = 0$; multiply by $4$ to get $3x^2 - 10x + 4 = 0$ ✓.

6. (Core) A quartic equation with real coefficients and leading coefficient $1$ has roots $2i$ and $3 - i$. Find the equation, and check your answer using the sum and product of its roots.

Solution

Real coefficients, so $-2i$ and $3 + i$ are also roots.

$$(x - 2i)(x + 2i) = x^2 + 4, \qquad (x - (3 - i))(x - (3 + i)) = (x - 3)^2 + 1 = x^2 - 6x + 10$$$$(x^2 + 4)(x^2 - 6x + 10) = x^4 - 6x^3 + 14x^2 - 24x + 40 = 0$$

Check. Sum of roots: $2i - 2i + (3 - i) + (3 + i) = 6$, and the formula gives $-\dfrac{-6}{1} = 6$ ✓. Product: $(2i)(-2i)(3 - i)(3 + i) = 4 \times 10 = 40$, and the formula gives $\dfrac{(-1)^4(40)}{1} = 40$ ✓.

7. (Core) One root of $2x^3 - 9x^2 + kx + 6 = 0$ is $x = 2$. Use the sum and product of the roots to find the other two roots, then find $k$.

Solution

Let the other roots be $\beta$ and $\gamma$.

Sum: $2 + \beta + \gamma = \dfrac{9}{2}$, so $\beta + \gamma = \dfrac{5}{2}$.

Product: $2\beta\gamma = \dfrac{(-1)^3(6)}{2} = -3$, so $\beta\gamma = -\dfrac{3}{2}$.

So $\beta$ and $\gamma$ are the roots of $x^2 - \dfrac{5}{2}x - \dfrac{3}{2} = 0$, or $2x^2 - 5x - 3 = 0$, which factors as $(2x + 1)(x - 3) = 0$. The other roots are $-\dfrac{1}{2}$ and $3$.

To find $k$, multiply out: $(x - 2)(2x^2 - 5x - 3) = 2x^3 - 9x^2 + 7x + 6$, so $k = 7$.

Check: substitute $x = 2$ into the original with $k = 7$: $16 - 36 + 14 + 6 = 0$ ✓.

8. (Challenge) The roots of $x^3 - 6x^2 + kx - 6 = 0$ form an arithmetic sequence. Find the roots and the value of $k$.

Solution

Write the roots as $a - d$, $a$, $a + d$.

Sum: $(a - d) + a + (a + d) = 3a = 6$, so $a = 2$.

Product: $(a - d)\,a\,(a + d) = a(a^2 - d^2) = \dfrac{(-1)^3(-6)}{1} = 6$. With $a = 2$: $2(4 - d^2) = 6$, so $d^2 = 1$ and $d = \pm 1$.

Either way the roots are $1$, $2$ and $3$. Then

$$(x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$$

so $k = 11$.

9. (Challenge) The roots of $x^3 - 2x^2 + 5x - 3 = 0$ are $\alpha$, $\beta$ and $\gamma$. Find a cubic equation whose roots are $\alpha + 1$, $\beta + 1$ and $\gamma + 1$, and check it using the sum and product of the roots.

Solution

Substitute $x - 1$ for $x$:

$$\begin{aligned} &(x - 1)^3 - 2(x - 1)^2 + 5(x - 1) - 3 \\ &= (x^3 - 3x^2 + 3x - 1) - (2x^2 - 4x + 2) + (5x - 5) - 3 \\ &= x^3 - 5x^2 + 12x - 11 \end{aligned}$$

The new equation is $x^3 - 5x^2 + 12x - 11 = 0$.

Check the sum: the old sum is $\alpha + \beta + \gamma = 2$, so the new sum should be $2 + 3 = 5$. The new equation gives $-\dfrac{-5}{1} = 5$ ✓.

Check the product: the new equation gives $\dfrac{(-1)^3(-11)}{1} = 11$. And $(\alpha + 1)(\beta + 1)(\gamma + 1)$ is the old cubic $p(x) = (x - \alpha)(x - \beta)(x - \gamma)$ evaluated at $x = -1$, times $-1$: $-p(-1) = -(-1 - 2 - 5 - 3) = 11$ ✓.