# Solving quadratic equation using formula

In this chapter we will learn to solve quadratic equation using formula with the help of solved example.

At the end of the chapter, we have provided solved problems for your practice.

There are basically two methods to solve quadratic equations;

(a) Factorization method
(b) Formula method

We have already discussed factorization method in previous chapter.

Here we will discuss the formula method.

### Solving quadratic equation using formula method

Let the given quadratic equation is;

a\mathtt{x^{2} +bx+c=0}

The solution of quadratic equation is given by formula;

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

By putting the relevant values in above expression, you will get solution of given quadratic equations.

### Types of solution of quadratic equation

There are two types of solution of quadratic equation;

(a) Real solution

(b) Imaginary solution

For given quadratic equation, the type of solution can be found by value of determinant.

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

If D > 0, then quadratic equation has real solution.

If D < 0, then quadratic equation has imaginary solution.

I hope you understood the theory. Let us solve some problems for better understanding.

## Quadratic Equation – Solved Problems

Example 01

\mathtt{2x^{2} -11x+14=0}

Solution
Let’s find the determinant first.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{D=( -\ 11)^{2} -4( 2)( 14)}\\\ \\ \mathtt{D=\ 121-112}\\\ \\ \mathtt{D\ =\ 9}

Since D > 0, we will get real solution.

Now let’s find the solution of quadratic equation using formula;

\mathtt{x=\frac{-b\pm \sqrt{b^{2} -4ac\ }}{2a}}\\\ \\ \mathtt{x=\ \frac{-( -11) \pm \sqrt{9}}{2( 2)}}\\\ \\ \mathtt{x=\ \frac{11\pm 3}{4}}\\\ \\ \mathtt{x=\frac{11+3}{4} \ \ \&\ x=\frac{11-3}{4}}\\\ \\ \mathtt{x=\frac{14}{3} \ \ \&\ x=\frac{8}{4}}\\\ \\ \mathtt{x=\frac{14}{3} \ \&\ x\ =\ 2}

Hence, x = 14/3 and x = 2 is the solution of given equation.

Example 02

\mathtt{4x^{2} +4x+1=0}

Solution
Check the value of determinant;

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{D=( 4)^{2} -4( 4)( 1)}\\\ \\ \mathtt{D=\ 16-16}\\\ \\ \mathtt{D\ =\ 0}

Since D = 0, the above quadratic equations have real solution.

\mathtt{x=\frac{-b\pm \sqrt{b^{2} -4ac\ }}{2a}}\\\ \\ \mathtt{x=\ \frac{-( 4) \pm \sqrt{0}}{2( 4)}}\\\ \\ \mathtt{x=\ \frac{-4}{8}}\\\ \\ \mathtt{x=\frac{-1}{2}}

Hence, x = -1/2 is the solution of given quadratic equation.

Example 03

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

Solution
Finding the determinants of the equation.

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

Since D < 0, we will get imaginary solution.

\mathtt{x=\frac{-b\pm \sqrt{b^{2} -4ac\ }}{2a}}\\\ \\ \mathtt{x=\ \frac{-( 4) \pm \sqrt{-4}}{2( 1)}}

Note that;
\mathtt{\sqrt{-1} =\ i}

Putting this value in main equation we get;

\mathtt{x=\ \frac{-4\pm 2i}{2}}\\\ \\ \mathtt{x=\frac{-4+2i}{2} \ \ \&\ x=\frac{-4-2i}{2}}\\\ \\ \mathtt{x=-2+i\ \ \&\ \ x=-2-i}

Hence, (-2 + i ) and (- 2 – i ) are the solution of given quadratic equation.

Note: Here ” i ” is called “iota” in math. This is a subject of higher grade mathematics, so we will deal with this later.

Example 04

\mathtt{x^{2} -7x-30=0}

Solution
Let’s find the determinant first.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{D=( -7)^{2} -4( 1)( -30)}\\\ \\ \mathtt{D=\ 49+120}\\\ \\ \mathtt{D\ =\ 169}

Since D > 0, the solution of quadratic equation is real.

\mathtt{x=\frac{-b\pm \sqrt{b^{2} -4ac\ }}{2a}}\\\ \\ \mathtt{x=\ \frac{-( -7) \pm \sqrt{169}}{2( 1)}}\\\ \\ \mathtt{x=\ \frac{7\pm 13}{2}}\\\ \\ \mathtt{x=\frac{7+13}{2} \ \ \&\ x=\frac{7-13}{2}}\\\ \\ \mathtt{x=10\ \ \&\ \ x=-3}

Hence, x = 10 and x = -3 are the solution of given equation.

Example 05

\mathtt{7x-1+x^{2} =0}

Solution
First arrange the quadratic equation from highest power to lowest power.

\mathtt{x^{2} +7x-1=0}

Calculate the determinant of given equation.

\mathtt{D\ =\ b^{2} -4ac}\\\ \\ \mathtt{D=( 7)^{2} -4( 1)( -1)}\\\ \\ \mathtt{D=\ 49+4}\\\ \\ \mathtt{D\ =\ 53}

Since D > 0, the quadratic equations have real solutions.