PX

^{2}+ QX + R = 0.

Where, P, Q and R are constants in equation. To solve a Quadratic Equation we have many methods like Factorization method in which quadratic equation is broken in two simpler equations. But not always the factorization technique works. We can use direct formula also:

X = [- Q ± √(Q

^{2}– 4PR)] / 2P,

When we solve quadratic equations, situation may arise where we need to solve Square roots to get final values. To know how to eliminate square roots from quadratic equation? Let us consider an example of it:

Suppose we have a simple quadratic equation: X2 – 625 = 0. To solve this equation we follow:

X

^{2}= 625 or X = √625,

Taking square root of any quantity gives us two possible results: positive and negative. So, above equation can be written as:

X = ± √625 = ± 25 i.e. X = + 25 or X = - 25,

Similarly if we consider more complex equation like: 4y

^{2 }+ 5y + 6 = 0, as it is not possible to factorize the following equation we need to use formula for solving quadratic equation:

X = [- Q ± √(Q

^{2}– 4PR)] / 2P,

Substituting the values of P, Q and R in the formula we get,

y = [- 5 ± √(5

^{2}– 4 * 4 * (6))] / 2 * 4,

This leaves us with two possible answers involving imaginary Numbers: y = -5 / 8 - 71i / 8 and y = -5 / 8 + 71i / 8. We can see here that two complex numbers are conjugate of each other. Thus we get two possible solutions by solving the square root.