# Solution Of Quadratic Equation In Complex Number System

## Quadratic equation in complex numbers

A complex number can be represented in the form of a+bi,

where a and b are real numbers and i is the imaginary number.

### Standard Form-

The standard form of quadratic equation is given by

 $ax^{2}+bx+c$

Where a,b,c are real numbers and $a\neq 0$.

The roots of the equation is given by-

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

Discriminant (D) $= \sqrt{b^{2}-4ac}$

The root can be of three form depending upon the value of D.

1. Distinct Roots- When D>0,then the quadratic equation has two distinct roots.

given by, $x_{1}= \frac{-b + \sqrt{b^{2}-4ac}}{2a},x_{2}= \frac{-b – \sqrt{b^{2}-4ac}}{2a}$

1. Equal Roots- When D=0,then the quadratic equation has two equal roots.

given by, $x_{1},x_{2}= \frac{-b}{2a}$

1. Complex Roots- When D<0, then the quadratic equation has two complex roots.

given by, $x_{1}= \frac{-b + i\sqrt{4ac – b^{2}}}{2a},x_{2}= \frac{-b – i\sqrt{4ac – b^{2}}}{2a}$

Note- A polynomial of degree n will have n roots. This is known as fundamental theorem of algebra.

The quadratic equation has degree of 2, thus they have 2 roots.

Power of i-

$i^{2}= -1$

$i^{3}= i.i^{2}= i(-1)=-i$

$i^{4}= i^{2}.i^{2}= (-1)(-1)=1$

Thus the general form is given by-

$i^{4k}= i$

$i^{4k+1}= -1$

$i^{4k+2}= -i$

$i^{4k+3}= 1$

 Example- Find the roots of the quadratic equation $x^{2}-x+1=0$. Solution- Comparing the given equation with the general form of the equation We have, a= 1, b= -1, c= 1 D= $b^{2}-4ac = (-1)^{2}-4(1)(1)=-3$ Thus the equation have two complex roots. $x= \frac{-b\pm \sqrt{D}}{2a}$ $x=\frac{-1\pm \sqrt{-3}}{2(1)}$ $x=\frac{-1\pm \sqrt{3}i}{2}$ Thus the roots are $x=\frac{-1 + \sqrt{3}i}{2}, \frac{-1 – \sqrt{3}i}{2}$ Example- Solve $\sqrt{5}x^{2}+x+\sqrt{5}= 0$ Solution- Comparing the given equation with the general form of the equation We have, $a = \sqrt{5},b = 1, c = \sqrt{5}$ Here Discriminant (D) = $b^{2} – 4ac = (1)^{2}- 4(\sqrt{5}).(\sqrt{5})= -19$ $x= \frac{-b\pm \sqrt{D}}{2a}$ $x= \frac{-1\pm \sqrt{-19}}{2\sqrt{5}}$ $x= \frac{-1\pm \sqrt{19}i}{2\sqrt{5}}$<

To learn more about NCERT Solutions for Complex Numbers and Quadratic equations click on this link, NCERT Solutions for Complex Numbers and Quadratic equations.

#### Practise This Question

Find the complex number z satisfying the equations z12z8i=53,  z4z8=1