Question

solve the quadratic equation 4x 2 +4bx-(a 2 -b 2 )=0

Answer 1

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

4x^2+4bx+b^2-a^2=0

(2x+b)^2-a^2=0

(2x+b-a)(2x+b+a)=0

x=(a-b)/2,-(a+b)/2

Answer 2

$\huge\mathbb{SOLUTION:-}$

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

$:\implies x {}^{2} + bx - \bigg( \frac{a {}^{2} - b {}^{2} }{4} \bigg) = 0$

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

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

$:\implies \bigg(x + \frac{b}{2} \bigg){}^{2} = \frac{a {}^{2} }{4}$

$:\implies x + \frac{b}{2} = \pm \frac{a}{2}$

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

$:\implies x = \frac{ - b \: - a}{2} , \frac{ - b + a}{2}$

$Hence, \: the \: roots \: are\implies \: -\bigg( \frac{a + b}{2} \bigg) \: and \: \bigg( \frac{a - b}{2} \bigg)$