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.

Quadratic Equation-

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

The locus of z satisfying the inequality log1/3|z+1|>log1/3|z1| is