Sum of the roots of a quadratic equation

For a quadratic equation of the form ax²+bx+c=0 with a non-negative discriminant Δ≥0, the sum of its two roots is equal to the negative ratio of the coefficient (b) of x to the coefficient (a) of x². In concise form: $$ x_1 + x_2 = - \frac{b}{a}$$

Why is this useful?

If one root of a quadratic equation is already known, the other can be found immediately without solving the equation again.

$$ x_1 = - \frac{b}{a} - x_2 $$

$$ x_2 = - \frac{b}{a} - x_1 $$

Example. Consider the quadratic equation $$ 2x^2+3x+1= 0 $$ If we know that one root is x₁=-1, we can determine the second root using the coefficients a=2 and b=3: $$ x_2 = - \frac{b}{a} - x_1 = - \frac{3}{2} - (-1) = \frac{3}{2} + 1 = \frac{-3+2}{2} = - \frac{1}{2} $$ So if one root is easy to identify at sight, there is no need to solve the quadratic formula again in order to obtain the second root.

A practical example

Consider the equation

$$ 2x^2+3x+1= 0 $$

The discriminant is Δ=1, which is positive, so the equation has two distinct real solutions.

The roots are:

$$ x = \frac{-3 \pm \sqrt{9-4(2)(1)}}{2 \cdot 2}$$

$$ x = \frac{-3 \pm \sqrt{9-8}}{4}$$

$$ x = \frac{-3 \pm \sqrt{1}}{4}$$

$$ x = \frac{-3 \pm 1}{4} $$

$$ x = \begin{cases} x_1 = \frac{-3-1}{4}=-1 \\ \\ \frac{-3+1}{4} = - \frac{1}{2}\end{cases} $$

Now add the two roots:

$$ x_1 + x_2 = -1 + ( - \frac{1}{2} ) $$

$$ x_1 + x_2 = -1 - \frac{1}{2} = \frac{-2-1}{2} = -\frac{3}{2} $$

Since the coefficients are a=2 and b=3, compute the signed ratio:

$$ - \frac{b}{a} = - \frac{3}{2} $$

The result is identical, as expected.

The proof

Consider the general quadratic equation:

$$ ax^2+bx+c= 0 $$

Because the discriminant satisfies Δ≥0, the equation has two real solutions, which may be equal or distinct.

Note. If the discriminant were negative, the radical would not yield any real solutions.

The two solutions are given by:

$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$

$$ x_1 = \frac{-b - \sqrt{b^2-4ac}}{2a} $$

$$ x_2 = \frac{-b + \sqrt{b^2-4ac}}{2a} $$

Now compute their sum x₁+x₂:

$$ x_1 + x_2 = \frac{-b - \sqrt{b^2-4ac}}{2a} + \frac{-b + \sqrt{b^2-4ac}}{2a} $$

$$ x_1 + x_2 = \frac{-b - \sqrt{b^2-4ac} - b + \sqrt{b^2-4ac}}{2a} $$

$$ x_1 + x_2 = \frac{-2b}{2a} $$

This shows that the sum of the two roots is simply the negative ratio of the coefficients b/a:

$$ x_1 + x_2 = - \frac{b}{a} $$

And the result is proved.