In this chapter, we will learn about different types of roots we get while solving quadratic equation.

After reading the chapter, you will learn to classify the type of root in the given quadratic equation using a shortcut formula.

At the end of the chapter, some solved problems are also given for your practice.

Before moving on with the main topic, let us review the basic concepts.

## Quadratic equation and its roots

What are quadratic equation ?

The equation that can be expressed in the form of \mathtt{ax^{2} +bx+c=0} are called quadratic equation.

### What are roots of quadratic equation ?

The roots of the quadratic equations are the point that satisfy the given equation.**For example;**

Consider the quadratic equation \mathtt{x^{2} +2x-3=0}

Here the points, x= 1 & -3 are the roots because if we put the values in the equation, the expression will be satisfied.

Let us put x = 1 in the given equation.

\mathtt{x^{2} +2x-3=0}\\\ \\ \mathtt{( 1)^{2} +2( 1) -3=0}\\\ \\ \mathtt{1+2-3=0}\\\ \\ \mathtt{3-3\ =\ 0}\\\ \\ \mathtt{0\ =\ 0}\\\ \\ \mathtt{LHS\ =\ RHS}

Since the equation is satisfied, it confirms that point x = 1 is the root.

If you want to read about **roots of quadratic equation** in detail, click the red link.**Note:**

In the quadratic equation, there can be only two roots possible.

## Identify type of roots for given quadratic equation

We already know that quadratic equations have two roots. To identify the type of roots, follow the below points.

Let the given quadratic equation is \mathtt{ax^{2} +bx+c=0} **(a) Determinant (D) = 0**

If, \mathtt{b^{2} -4ac\ =\ 0} ; then the quadratic equations have **real and equal roots**.

The roots of quadratic equation is given as;

\mathtt{x=\frac{-b\pm \sqrt{b^{2} -4ac}}{2a}}\\\ \\ \mathtt{x=\frac{-b\pm \sqrt{0}}{2a}}\\\ \\ \mathtt{x=\frac{-b}{2a}}

Hence, values of both roots are -b/2a.

**(b) Determinant (D) > 0**

If, \mathtt{b^{2} -4ac\ >\ 0} , then we have **real and different roots.**

The value of roots is given as;

\mathtt{x_{1} =\frac{-b+\sqrt{b^{2} -4ac}}{2a}}\\\ \\ \mathtt{x_{2} =\frac{-b-\sqrt{b^{2} -4ac}}{2a}}

Hence, using above two formulas will give you value of roots for given quadratic equation.

**(c) Determinant D < 0**

When D < 0 ,we will get roots which are **different and imaginary** in nature.

The roots can be expressed as;

\mathtt{x =\frac{-b\pm \ imaginary\ number}{2a}}

I hope you understood the above three points. Let us solve some problems for further clarity.

## Types of Quadratic equation roots – Solved examples

**Example 01**

Check if the below quadratic equation have real or imaginary roots.

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

**Solution**

Check the determinant of given equation.

\mathtt{D\ =\ b^{2} -4ac}

Putting the values;

\mathtt{\Longrightarrow \ 5^{2} -4( 1)( 6)}\\\ \\ \mathtt{\Longrightarrow \ 25\ -\ 24}\\\ \\ \mathtt{\Longrightarrow \ 1}

Since D > 0, the quadratic equation has real and different roots.

**Example 02**

Check if the equation have real or imaginary roots.

\mathtt{x^{2} +4x+5=0}

**Solution**

Finding the value of determinant.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{\Longrightarrow \ 4^{2} -4( 1)( 5)}\\\ \\ \mathtt{\Longrightarrow \ 16\ -\ 20}\\\ \\ \mathtt{\Longrightarrow \ -4}

Since D < 0, the quadratic equation has imaginary roots.

**Example 03**

The below quadratic equation has real roots. Find the value of k.

\mathtt{x^{2} -kx+25=0}

**Solution**

We know that quadratic equation with equal roots have determinant = 0.

\mathtt{\ b^{2} -4ac=0}\\\ \\ \mathtt{\ ( -k)^{2} -4( 1)( 25) =0}\\\ \\ \mathtt{\ k^{2} \ -\ 100=0}\\\ \\ \mathtt{k^{2} =\ 100}\\\ \\ \mathtt{k\ =\ 10}

Hence, quadratic equation becomes \mathtt{x^{2} -10x+25=0}

**Example 04**

The given quadratic equation has real & equal roots. Find the value of k.

\mathtt{x^{2} -22x+( k+3) =0}

**Solution**

When roots are real and equal, determinant is zero.

\mathtt{\ b^{2} -4ac=0}\\\ \\ \mathtt{\ ( -22)^{2} -4( 1)( k+3) =0}\\\ \\ \mathtt{484-4k-12=0}\\\ \\ \mathtt{472-4k=0}\\\ \\ \mathtt{k=\ \frac{472}{4}}\\\ \\ \mathtt{k\ =\ 118}

Hence, the value of k is 118

**Next chapter** : **Remainder theorem of polynomials**