# How to Find Complex Roots of a Quadratic Equation?

An equation of the form ax2 + bx + c = 0 is called a quadratic equation, where a, b, and c are real numbers and a ≠ 0. A quadratic equation has complex roots if its discriminant is less than zero. We can find the square root of negative real numbers in the set of complex numbers. A number that can be expressed in the form a + ib is referred to as a complex number. Here a and b are real numbers. The value of i equals √-1.

Complex numbers and quadratic equations are one of the most important chapters in the preparation of competitive entrance exams. Students can expect 1-3 questions from this topic for JEE Main and other exams. In this article, we will learn how to solve complex quadratic equations.

## Methods to Find Complex Roots of a Quadratic Equation

1. The solutions to the above equation are available in the set of complex numbers which are given by

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

2. If the equation is of the form z2 = (x + iy)2, expand the expression and equate real part and imaginary part. Then solve for x and y.

## Solved Examples

Let us have a look at some examples.

Example 1:

Solve the equation 2x2 + x + 1 = 0

Solution:

Given 2x2 + x + 1 = 0

b2 – 4ac = 1 – 4×2×1

= 1 – 8

= -7

-7 < 0

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

So x = ( – 1 ± √7 i )/4

Hence the roots are ( – 1 + √7 i )/4 and ( – 1 – √7 i )/4.

Example 2:

Solve x2 + 3x + 5 = 0

Solution:

Given x2 + 3x + 5 = 0

b2 – 4ac = 32 – 4×1×5

= 9 – 20

= -11

< 0

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

x = (- 3 ± √11 i)/2

Hence the roots are ( – 3 + √11 i )/2 and ( – 3 – √11 i )/2.

Example 3.

Solve z2 = (x + iy)2 = 15 + 8i

Solution:

z2 = (x + iy)2 = 15 + 8i

(x + iy)2 = (x2 – y2) + 2xyi = 15 + 8i

Comparing real and imaginary part

x2 – y2 = 15 …(i)

2xyi = 8i

xy = 8/2 = 4

The factors of 4 are (4,1) (-4,-1), (2,2) and (-2,-2).

Substitute these values in (i) and check which value satisfies the equation.

We get (4,1) and (-4, -1) satisfies (i).

Hence, z = 4 + i or -4 – i are the roots of the equation.

Example 4.

Find the roots of x2+4x+5 = 0.

Solution:

Given x2+4x+5 = 0.

b2-4ac = 16-4(1)(5)

= -4<0

x = (-4±√-4)/2(1)

= (-4±2i)/2

= -2±i