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

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

where x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0. The sum of roots x1 + x2 and the product of roots x1·x2 are common to problems involving quadratic equation.

Derivation of the Sum of Roots
$x_1 + x_2 = \dfrac{-b + \sqrt{b^2-4ac}}{2a} + \dfrac{-b - \sqrt{b^2-4ac}}{2a}$

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

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

$x_1 + x_2 = - \, \dfrac{b}{a}$

Derivation of the Product of Roots
$x_1 \, x_2 = \left( \dfrac{-b + \sqrt{b^2-4ac}}{2a} \right) \left( \dfrac{-b - \sqrt{b^2-4ac}}{2a} \right)$

By difference of two squares:
$x_1 \, x_2 = \dfrac{b^2 - (b^2-4ac)}{4a^2}$

$x_1 \, x_2 = \dfrac{4ac}{4a^2}$

$x_1 \, x_2 = \dfrac{c}{a}$

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