If a, b, c are non-zero real numbers such that a + b + c = 0, then what are the roots of the equation ax² + bx + c = 0 ? 

Formula used:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Solution of quadratic equation ax² + bx + c = 0 is given by

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given that 

a + b + c = 0

b = - (a + c)              ------(1)

Root of quadratic equation ax2 + bx + c = 0  are

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {(a+c) \pm \sqrt{[-(a+c)]^2-4ac} \over 2a}\)

\(x = {(a+c) \pm \sqrt{a^2+2ac+c^2-4ac} \over 2a}\)

\(x = {(a+c) \pm \sqrt{a^2-2ac+c^2} \over 2a}\)

\(x = {(a+c) \pm \sqrt{(a-c)^2} \over 2a}\)

\(x = {(a+c) \pm {(a-c)} \over 2a}\)

Considering +ve & -ve signs alternatively,

\(x = {(a+c) +{(a-c)} \over 2a}\) & \(x = {(a+c) -{(a-c)} \over 2a}\) 

\(x = \frac{{2a}}{ 2a}\) & \(x = \frac{{2c}}{ 2a}\)

 \(x = 1, \ \ x = \frac{c}{a}\)