Solution Of Quadratic Equation In Complex Number System

In Algebra, we will find certain quadratic equations with negative roots in their solutions. These kinds of roots have imaginary numbers and the roots are sometimes called the complex roots. In this article, we will discuss the complex numbers and quadratic equations and the nature of roots in detail.

Complex Numbers and Quadratic Equations

A complex number can be represented in the form of a+bi, which is the combination of both the real numbers and the imaginary numbers. Here, a and b are real numbers and i is the imaginary number.

Solution of Complex Quadratic Equations

A quadratic equation is an equation, where atleast one term should be squared. The maximum degree of the equation must be two. For example, 5x2+3x+3 =0. In this case, the highest order of the equation is 2. So, the given equation is a quadratic equation.

Standard Form

The standard form of the quadratic equation is given by


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

Nature of the Roots

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 a 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-

\(i^{4k}= i\)

\(i^{4k+1}= -1\)

\(i^{4k+2}= -i\)

\(i^{4k+3}= 1\) 

For Complex Case

As we know, the discriminant of the quadratic equation is given by;

D = √(b2-4ac)

If 4ac > b,then the solution will have non-zero imaginary part. Therefore, we can write;

\(\sqrt{\left(-\left(4 a c-b^{2}\right)\right)}=\sqrt{\left(4 a c-b^{2}\right)} \cdot \mathrm{i}\)

Now, the square root \(\sqrt{\left(4 a c-b^{2}\right)}\) is positive and real. Therefore, for complex number case, we get the solution for quadratic equation as;

\(\mathrm{x}=\frac{-b \pm i \cdot\left(4 a c-b^{2}\right)^{1 / 2}}{2 a}\)

Solved Examples

Example 1: Find the roots of the quadratic equation \(x^{2}-x+1=0\).


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 2: Solve \(\sqrt{5}x^{2}+x+\sqrt{5}= 0\)


Comparing the given equation with the general form of the equation

We have, \(a = \sqrt{5},b = 1, c = \sqrt{5}\)

Here Discriminant (D) = \(b^{2} – 4ac = (1)^{2}- 4(\sqrt{5}).(\sqrt{5})= -19\)

\(x= \frac{-b\pm \sqrt{D}}{2a}\)

\(x= \frac{-1\pm \sqrt{-19}}{2\sqrt{5}}\)

\(x= \frac{-1\pm \sqrt{19}i}{2\sqrt{5}}\)

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