Question

Solve the following quadratic equation for x 4 x square + 4 x minus (a square minus b square) is equal to zero

Answer 1

Answer:

I hope this is right answer

Answer 2

Concept:

We first recall the concept of quadratic equations to solve the given problem.

An equation of degree 2 of the form $Ax^{2} +Bx+C=0$             (1)              is called the quadratic equation.                      

where A , B and C ∈ $R$

The solution of Quadratic equation is obtained by using the quadratic formula x = $\frac{-B\pm \sqrt{D}}{2A}$ where D is discriminant given by D = $B^{2} -4 A C$

Given :

The quadratic equation is:

$4x^{2} +4bx-(a^{2}-b^{2})=0$                                                               (2)

To find:

The solution of quadratic equation.

Solution:

On comparing equation (2) with equation (1) we get,

A=4 , B=4b and C = $-(a^{2}-b^{2})$

Discriminant, D = $(4b)^{2} -4 (4)(-(a^{2}-b^{2}))$

                     D = $16b^{2} +16a^{2} -16b^{2}$

                     D = $16a^{2}$

Solution of Quadratic Equation is :

                     $x$ = $\frac{-4b\pm\sqrt{16a^{2} } }{2 (4)}$

                     $x=\frac{-4b\pm 4a} {8}$

                    $x=\frac{ -b\pm a}{2}$

Hence, the solution is given by

$x_{1}=\frac{-b+a}{2}\; and\; x_{2} =\frac{-b-a}{2}$  where $x_{1} and\; x_{2}$ are roots of quadratic equation.