The nature of roots of quadratic equations is determined by whether the roots are real, complex, they are equal, or unequal. The nature of the roots of a quadratic equation can be determined by the **discriminant** of the equation. Let’s learn about that in this post.

**What are Roots of a Quadratic Equation**

An equation of degree 2 is called a quadratic equation. The general form of a quadratic equation is as follows:

ax^{2}+bx+c = 0 …(1)

where a ≠ 0. Here a, b, and c are real numbers.

The value of x satisfying the above equation (1) is called the roots of the quadratic equation (1).

**Methods of Finding the Nature of Roots of Quadratic Equations**

Note that there are two roots of a quadratic equation, and they can be determined by Shreedhara Acharya’s formula which are given below:

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

Write D=b^{2}-4ac.

This quantity D=b^{2}-4ac is called the **discriminant** of the given quadratic equation.

Then, x = (-b+√D)/2a and (-b-√D)/2a.

Depending upon the discriminant D, the nature of the roots of the quadratic equation (1) can be determined as follows:

**Real and Unequal Roots**: If D is greater than 0, then both the roots will be real and unequal. If a, b, and c are rational numbers and D is a perfect square then the roots will be rational numbers; otherwise, they are irrational.**Real and Equal Roots**: If D is equal to 0, then both the roots will be real and equal and the roots are x=-b, -b.**Complex Roots**: If D is less than 0, then both the roots will be complex numbers. The roots will be conjugated to each other, and they are x = (-b+i√-D)/2a and (-b-i√-D)/2a.

The above discussion about the nature of the roots of a quadratic equation can be summarized in the table below:

D>0 | Real and unequal roots |

D=0 | Real and equal roots |

D<0 | Complex roots |

Let us now learn the above method with examples to find the nature of the roots of a quadratic equation.

**Solved Examples**

**Example 1**: Find the nature of the roots of x^{2 }+2x+5=0.

**Solution:**

Comparing the equation x^{2 }+2x+5=0 with ax^{2 }+bx+c=0, we get that

a=1, b=2 and c=5

Then the discriminant D=b^{2}-4ac = 2^{2} -4×1×5 = 4-20 = -18<0.

As D<0, the above method says that the roots of x^{2 }+2x+5=0 are complex numbers (conjugate to each other).

**Have You Read These:**

How to find roots of a quadratic equation

**FAQs**

### Q1: How to find the nature of the roots of a quadratic equation ax^{2 }+bx+c=0.

**Answer**: The discriminant D of the equation ax^{2 }+bx+c=0 is equal to D=b^{2}-4ac. If D=0 then the equation has real and equal roots, if D>0 then the equation has two real and distinct roots, if D<0 then it has two complex roots conjugated to each other.