The Sum and Product of the Roots

A quadratic built from its roots

The solutions of a quadratic equation are also called its roots. If the roots are 2 and 3, then (x − 2)(x − 3) = 0 is an equation with exactly those roots, by the zero product property.

Call the two roots $\alpha$α and $\beta$β, the Greek letters alpha and beta. Then $\left(x-\alpha \right)\left(x-\beta \right)=0$(x − α)(x − β) = 0 has the roots $\alpha$α and $\beta$β. Multiplying the left side by any number a other than 0 does not change where it is zero, so every quadratic with the roots $\alpha$α and $\beta$β can be written as $a\left(x-\alpha \right)\left(x-\beta \right)$a(x − α)(x − β), where a is the coefficient of $x^2$x².

Expand the brackets: $\left(x-\alpha \right)\left(x-\beta \right)=x^2-\alpha x-\beta x+\alpha \beta =x^2-\left(\alpha +\beta \right)x+\alpha \beta$(x − α)(x − β) = x² − αx − βx + αβ = x² − (α + β)x + αβ. Multiplying through by a gives $a{x}^2-a\left(\alpha +\beta \right)x+a\alpha \beta$ax² − a(α + β)x + aαβ.

Matching the coefficients

The same quadratic is also written $a{x}^2+bx+c$ax² + bx + c. Two ways of writing one quadratic must have the same coefficients, term by term, as in Finding Unknown Coefficients by Matching.

Match the x terms: $b=-a\left(\alpha +\beta \right)$b = −a(α + β). Divide both sides by −a, and the sum of the roots is $\alpha +\beta =-\frac{b}{a}$α + β = −b/a.

Match the constant terms: $c=a\alpha \beta$c = aαβ. Divide both sides by a, and the product of the roots is $\alpha \beta =\frac{c}{a}$αβ = c/a.

The curve is $y=\left(x-r_1\right)\left(x-r_2\right)$y = (x − r₁)(x − r₂), with its two roots as the handles. The expansion above it changes as they move: the sum of the roots appears, with a minus sign, as the coefficient of x, and their product appears as the constant term.

Reading them off without solving

Take $2x^2-10x+3=0$2x² − 10x + 3 = 0. Here a = 2, b = −10 and c = 3. The sum of the roots is $-\frac{b}{a}=\frac{10}{2}=5$−b/a = 10/2 = 5, and their product is $\frac{c}{a}=\frac{3}{2}$c/a = 3/2.

Neither root was found. The quadratic formula gives them as $\frac{\left(5+\sqrt{19}\right)}{2}$(5 + √19)/2 and $\frac{\left(5-\sqrt{19}\right)}{2}$(5 − √19)/2, which are awkward to work with, but their sum and product are simple numbers that can be read straight off the equation.

Checking on a quadratic that factors

$x^2-5x+6$x² − 5x + 6 factors as (x − 2)(x − 3), so its roots are 2 and 3. Their sum is 2 + 3 = 5, and $-\frac{b}{a}=-\frac{\left(-5\right)}{1}=5$−b/a = −(−5)/1 = 5. Their product is 2 × 3 = 6, and $\frac{c}{a}=\frac{6}{1}=6$c/a = 6/1 = 6. Both rules agree.

The rules also work when a is not 1. $2x^2-7x+3$2x² − 7x + 3 factors as (2x − 1)(x − 3), so its roots are $\frac{1}{2}$1/2 and 3. The rules give the sum $-\frac{b}{a}=\frac{7}{2}$−b/a = 7/2 and the product $\frac{c}{a}=\frac{3}{2}$c/a = 3/2, and indeed $\frac{1}{2}+3=\frac{7}{2}$1/2 + 3 = 7/2 and $\frac{1}{2}\times 3=\frac{3}{2}$1/2 × 3 = 3/2.

Watch the signs

The sum has a minus sign in front of b, and the product has no minus sign. For $x^2+5x+6=0$x² + 5x + 6 = 0, the roots add to −5 and multiply to 6. A positive product means the roots have the same sign, and a negative sum means that sign is negative. The roots are −2 and −3: −2 + (−3) = −5 and −2 × (−3) = 6.

Building an equation from its roots

The rules also run backward. With a = 1, the expansion $x^2-\left(\alpha +\beta \right)x+\alpha \beta$x² − (α + β)x + αβ says that a quadratic equation with given roots is $x^2-($x² − (sum of the roots)x + (product of the roots) = 0.

For the roots 2 and −5, the sum is −3 and the product is −10, so the equation is $x^2-\left(-3\right)x+\left(-10\right)=0$x² − (−3)x + (−10) = 0, which is $x^2+3x-10=0$x² + 3x − 10 = 0. Check by factoring: $x^2+3x-10=\left(x-2\right)\left(x+5\right)$x² + 3x − 10 = (x − 2)(x + 5), which is zero at x = 2 and x = −5.

The usual mistakes

Leaving out the minus sign in the sum. For $2x^2-10x+3$2x² − 10x + 3, the sum is $-\frac{\left(-10\right)}{2}=5$−(−10)/2 = 5, not −5.

Forgetting to divide by a. The sum for $2x^2-10x+3$2x² − 10x + 3 is $\frac{10}{2}=5$10/2 = 5, not 10, and the product is $\frac{3}{2}$3/2, not 3.

Mixing up the two rules. The sum is built from b and the product from c.