In this chapter, we will learn the concept of roots of quadratic equation with solved examples.

At the end, we have also provided some solved examples for practicing the concept.

## What are roots of quadratic equation ?

Consider quadratic equation \mathtt{ax^{2} +bx+c} .

The **value of x which satisfy the above quadratic equation to zero** are called **roots of quadratic equation**.

**For example;**

For the quadratic equation \mathtt{x^{2} +2x-63=0} , the two roots are -9 and 7.

It means that when we put the value of x = -9 or 7 in the above quadratic expression, the equation will be satisfied.

**Putting x = -9**

\mathtt{x^{2} +2x-63=0}\\\ \\ \mathtt{( -9)^{2} +2( -9) -63=0}\\\ \\ \mathtt{81\ -\ 18\ -63=0}\\\ \\ \mathtt{0\ =\ 0}\\\ \\ \mathtt{L.H.S=\ R.H.S}

Hence putting x = -9, the equation is satisfied.

We get similar result, when we put x = 7 in the above equation.

Let us now try **put x = 3** in the same equation.

\mathtt{x^{2} +2x-63=0}\\\ \\ \mathtt{( 3)^{2} +2( 3) -63=0}\\\ \\ \mathtt{9\ +6\ -63=0}\\\ \\ \mathtt{-48=\ 0}\\\ \\ \mathtt{L.H.S\neq \ R.H.S}

While putting x = 3, the equation is not satisfied.

Hence, **3 is not the root of given quadratic equation.**

## Showing Quadratic roots in the graph

We know that in graph, the quadratic equation is expressed through parabola.

Here, the** roots are two points at which the parabola intersect the x axis where value of y is zero**.

For example;

Given below is the graph of quadratic equation \mathtt{4x^{2} +15x+10}

Note the two points A & B where the parabola intersect the x axis.

These two points are the roots of quadratic equation.

Note that in quadratic equation, there are only two roots. Also, these roots can be real or imaginary numbers depending on the given quadratic equation.

### How to find roots of quadratic equation ?

The roots of quadratic equation can be easily found by factorizing the given expression.

We have already discussed different methods of factorization. Given below are the relevant links for your reference.

**Factorization of quadratic equation**

**Quadratic equation factorization using formula**

I hope you understood the concepts. Let us solve some problems for further clarity.

## Roots of Quadratic equation – Solved problems

**Example 01**

Find the roots of below quadratic equation.

\mathtt{x^{2} +5x+6=0} **Solution**

Factorizing the quadratic equation using simple method.

\mathtt{x^{2} +5x+6=0}\\\ \\ \mathtt{x^{2} +2x+3x+6=0}\\\ \\ \mathtt{x( x+2) +3( x+2) =0}\\\ \\ \mathtt{( x+3)( x+2) =\ 0}\\\ \\ \mathtt{x+3=0\ \ or\ \ x+2=0}\\\ \\ \mathtt{x=\ -3\ \ \ or\ x\ =\ -2\ }

Hence, -3 & -2 are the roots of quadratic equation.

**Example 02**

For the quadratic equation \mathtt{x^{2} -15x+36=0} , check if numbers 12 and 4 are the roots or not.

**Solution**

We know that roots of the quadratic equation, satisfy the expression.**Checking for number 12**.

\mathtt{x^{2} -15x+36=0}\\\ \\ \mathtt{( 12)^{2} -15( 12) +36=\ 0}\\\ \\ \mathtt{144-\ 180\ +\ 36\ =\ 0}\\\ \\ \mathtt{0\ =\ 0\ }

Since, the equation is satisfied, number 12 is the root of given quadratic equation.

**Checking for number 4**.

\mathtt{x^{2} -15x+36=0}\\\ \\ \mathtt{( 4)^{2} -15( 4) +36=\ 0}\\\ \\ \mathtt{16-\ 60\ +\ 36\ =\ 0}\\\ \\ \mathtt{-8=\ 0}\\\ \\ \mathtt{LHS\neq RHS}

Since, the equation is not satisfied, number 4 is not the root of given quadratic equation.

**Example 03**

For the given quadratic equation \mathtt{x^{2} +2x+k=0} . If one of the root of quadratic equation is 11, then find value of k.

**Solution**

We know that root of quadratic equation satisfy the expression.

**Putting x = 11** in the equation;

\mathtt{x^{2} +2x+k=0}\\\ \\ \mathtt{( 11)^{2} +2( 11) +k=0}\\\ \\ \mathtt{121\ +\ 22\ +\ k\ =\ 0}\\\ \\ \mathtt{143\ +\ k\ =\ 0}\\\ \\ \mathtt{k\ =\ -143}

Hence, the equation becomes \mathtt{x^{2} +2x-143=0}

**Example 04**

For the quadratic equation \mathtt{2x^{2} -8x-24=0} . Check if -2 and -3 are the roots.

**Solution**

We know that roots of quadratic equation satisfy the equation to zero.

**Put x = -2 in the equation;**

\mathtt{2x^{2} -8x-24=0}\\\ \\ \mathtt{2\ ( -2)^{2} -8( -2) -24=0}\\\ \\ \mathtt{2( 4) +16-24=0}\\\ \\ \mathtt{8\ +\ 16\ -\ 24\ =0}\\\ \\ \mathtt{0\ =\ \ 0}

Since LHS = RHS, **the number -2 is the root of quadratic equation**.

**Now put x = -3;**

\mathtt{2x^{2} -8x-24=0}\\\ \\ \mathtt{2\ ( -3)^{2} -8( -3) -24=0}\\\ \\ \mathtt{2( 9) +24-24=0}\\\ \\ \mathtt{18\ +\ 24\ -\ 24\ =0}\\\ \\ \mathtt{18\ =\ \ 0}\\\ \\ \mathtt{LHS\ \neq \ RHS}

Since, **-3 cannot satisfy the equation, hence it is not the root.**

**Next chapter** : **Types of roots of quadratic equation **