Question

Solve by using the method of completing the square :-


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

Answer 1

Given equation : 4x^2 + 4bx - ( a^2 - b^2 ) = 0

⇒ 4x^2 + 4bx = ( a^2 - b^2 )

Step 1 : Dividing both sides by the co efficient of x^2.

$\implies \dfrac{4x^2}{4} + \dfrac{4bx}{4} = \dfrac{(a^2-b^2)}{4}$

$\implies x^2 + bx = \dfrac{a^2-b^2}{4}$

Step 2 : Adding both sides the square of product of 1 / 2 and the co efficient of x.

$\implies x^2 + bx + \bigg( \dfrac{b}{2} \bigg)^2 = \dfrac{a^2-b^2}{4} + \bigg(\dfrac{b}{2} \bigg)^2$

From the identities of factorization, we know, a^2 - 2ab + b^2 = ( a - b )^2.

$\implies ( x + \dfrac{b}{2} )^2 = \dfrac{a^2-b^2+b^2}{2^2}$

$\implies ( x + \dfrac{b}{2} )^2 = \bigg( \dfrac{a^2}{2^2}\bigg)$

$\implies x + \dfrac{b}{2} = \dfrac{a}{2} \;\;Or\;\; -\dfrac{a}{2}$

$implies x = \dfrac{a}{2} - \dfrac{b}{2} \:\:Or\:\: - \dfrac{a}{2} - \dfrac{b}{2}$

$\implies x = \dfrac{a-b}{2}\;\;Or\;\; -\dfrac{(a+b)}{2}$

Therefore the value of x satisfying the given equation is ( a - b ) / 2  or  - ( a + b ) / 2

Answer 2

Hello!!

Question:-

Solve by using the method of completing the square :-

4x^2 + 4bx – (a^2 – b^2) = 0

Solution:

Given that,

=> 4x^2 + 4bx – (a^2 – b^2) = 0

=> 4x^2 + 4bx – a^2 + b^2 = 0

=> (2x)^2 + 2.(2x).b + b^2 – a^2 = 0

=> (2x + b)^2 – a^2 = 0

Now, Use given formula:

● a^2 – b^2 = (a – b)(a + b)

=>{2x + b – a}{2x + b + a} = 0

Hence, x = (a – b) / 2

OR,

x =  – (a + b) / 2

Hope It Helps!!!!!!