Product of the roots of a quadratic equation

For a quadratic equation of the form ax²+bx+c, with a non negative discriminant Δ≥0, the product of its two roots is equal to the ratio between the constant term c and the leading coefficient a. $$ x_1 \cdot x_2 = \frac{c}{a} $$

Why is this relation useful?

If I already know one root of the quadratic equation, I can immediately determine the other without solving the equation again.

$$ x_1 = \frac{c}{a \cdot x_2} $$

$$ x_2 = \frac{c}{a \cdot x_1} $$

Example. Consider the quadratic equation $$ x^2 +4x + 3 = 0 $$ If one root is known, for instance x₁=-1, I can determine the second root x₂ using the coefficients a=1 and c=3. $$ x_2 = \frac{c}{a \cdot x_1} = \frac{3}{1 \cdot (-1)} = -3 $$ This relationship is especially convenient when one root is easy to spot, or when a root is already known for another reason.

A practical example

Consider the quadratic equation

$$ x^2 +4x + 3 = 0 $$

The discriminant Δ is non negative

$$ \Delta = b^2-4ac = 4^2-4(1)(3) = 16 - 12 = 4 $$

Therefore, the product of the two roots x₁·x₂ is equal to c/a

$$ x_1 \cdot x_2 = \frac{c}{a} $$

Since c=3 and a=1

$$ x_1 \cdot x_2 = \frac{3}{1} = 3 $$

Check. Now I compute the roots of the equation: $$ x = \frac{-4 \pm \sqrt{4^2-4(1)(3)}}{2} $$ $$ x = \frac{-4 \pm \sqrt{16-12}}{2} $$ $$ x = \frac{-4 \pm \sqrt{4}}{2} $$ $$ x = \frac{-4 \pm 2}{2} $$ $$ x = \begin{cases} x_1 = \frac{-4-2}{2} = -3 \\ \\ x_2 = \frac{-4+2}{2} = -1 \end{cases} $$ Therefore, since x₁=-3 and x₂=-1, the product x₁·x₂ is equal to 3 $$ x_1 \cdot x_2 = (-3) \cdot (-1) = 3 $$ The final result matches the ratio c/a for the quadratic equation $$ x_1 \cdot x_2 = \frac{c}{a} = \frac{3}{1} = 3 $$ Consequently, if I know x₁=-3, I can determine x₂ without solving the equation again $$ x_2 = \frac{c}{a \cdot x_1} = \frac{3}{-3} = -1 $$ Likewise, if I know x₂=-1, I can determine x₁ $$ x_1 = \frac{c}{a \cdot x_2} = \frac{3}{-1} = - 3$$

The proof

Consider the general quadratic equation

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

By assumption, the discriminant Δ≥0 is non negative

$$ \Delta = b^2 - 4ac \ge 0 $$

Therefore, the equation has two distinct real roots x₁ and x₂

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

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

Now I compute the product of the two roots x₁·x₂

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

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

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

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

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

$$ x_1 \cdot x_2 = \frac{4ac }{4a^2} $$

$$ x_1 \cdot x_2 = \frac{c }{a} $$

The proof is complete

And so on.