Question

Solve quadratic equation by Factorization:


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

Answer 1

Hey there !!


▶ The given quadratic equation :-

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

=> 4x² + 4bx = ( a² - b² ) .

[Adding b² on both side ]

=> 4x² + 4bx + b² = a² - b² + b² .

=> (2x)² + 2 × 2x × b + b² = a² .

=> ( 2x + b )² = a² .

[ Taking square root on both side, we get ] .

=> 2x + b = ± a.

=> 2x + b = a or 2x + b = -a .

=> 2x = a - b or 2x = - a - b .

$\boxed{ \bf \therefore x = \frac{a - b}{2} \: \: \: or \: \: \: x = - \frac{a + b}{2} . }$


✔✔ Hence, it is solved ✅✅.



THANKS



#BeBrainly.

Answer 2

$\huge\mathbb{SOLUTION:-}$

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

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

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

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

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

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

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

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

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