The quadratic formula is a method that is used to find the roots of a quadratic equation. Several methods have been proposed for the solution of quadratic equations.

For a general quadratic equation of the form:

`ax^2 + bx + c = 0`

where a,b,c are constants (can be -ve) and where a !=0, the **quadratic formula** is given by

` x=(-b+-sqrt(b^2-4ac))/(2a)`

This formula is so important that you must learn it by heart for good

According to this solution, there are two **roots** of the quadratic equation. And they are given as:

`x=(-b+sqrt(b^2-4ac))/(2a) ` , ` x=(-b- sqrt(b^2-4ac))/(2a) `

i.e. one with the positive sign, while the other has a negative sign.

In the expression for the roots of a quadratic equation, the term present inside the square root `(b^2-4ac)` is termed as **discriminant** and
the value of this expression is of great significance.

There are 3 different possibilities with regards to the value of discriminant and the consequences of each type of value have been discussed below:

If the discriminant has a **positive** value, the two roots of the quadratic equation are both **distinct and real**.

If the discriminant is** equal to zero,** the two roots are** equal and real.**

The third possibility arises when the discriminant is **negative** i.e. less than zero. In this case, the two roots are **distinct and conjugate complex number.**

In order to solve this equation using quadratic formula discussed above, we need to compare the coefficients of the given equation with the general form of quadratic equation:

`ax^2 + bx + c = 0`

Comparison shows that

a = 1

b = - 5

c = 3

Using the formula, we have

` x=(-b+-sqrt(b^2-4ac))/(2a) `

=>` x=(-(-5) +- sqrt((-5)^2-4(1)(3)))/(2(1))`

=> ` x=( +5 +- sqrt(25 - 12 ))/2`

=> ` x=( +5 +- sqrt(13 ))/2`

Where the two roots are real and distinct ( 13 > 0 )

Compare the given equation with the general form of quadratic equation

`ax^2 + bx + c = 0`

Comparison shows that

a = 4

b = - 6

c = 2

Use the quadratic formula to find x.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-6)+-sqrt([(-6)]^2-4(4)(2)))/(2(4))`

`x=( 6 +-sqrt( 36-32 ))/8`

`x=( 6 +-sqrt( 4 ))/8`

`x=( 6 +-2 )/8`

=> `x=( 6 + 2 )/8 ` and `x=( 6 - 2 )/8`

=> x=1 and ` x=( 4 )/8 =( 1 )/2 `

Compare the given equation with the general form of quadratic equation

`ax^2 + bx + c = 0 `

But wait a minute, where is c? What is different in this quadratic equation? The right hand-side of this equation is non-zero.

We can subtract 11 from both sides of the equation to get

`x^2 + 4x - 11 = 11 - 11`

=> ` x^2 + 4x - 11 = 0`

Comparison shows that

a = 1

b = 4

c = - 11

Use the quadratic formula to find x.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-4+-sqrt((4)^2-4(1)(-11)))/(2(1))`

`x=(-4+-sqrt( 16+44 ))/2`

`x=(-4+-sqrt( 60 ))/2`

`x=(-4+-2sqrt( 15 ))/2`

`x= -2+-sqrt( 15 ) `

Compare the given equation with the general form of quadratic equation

`ax^2 + bx + c = 0`

Comparison shows that

a = 1

b = 4

c = 4

Use the quadratic formula to find x.

`x=(-b+-sqrt(b^2-4ac))/2a`

`x=(-4+-sqrt((4)^2-4(1)(4)))/(2(1))`

`x=(-4+-sqrt( 16-16 ))/2`

`x=(-4+-0 )/2`

`x= -2,-2`

Hence the two roots are real and duplicate since `b^2 - 4ac = 0 `

© 2022 iPracticeMath | All Rights Reserved | Terms of Use.