Question

4x^2 + 4bx -(a^2- b^2)= 0. Solve by completing the square method.

Answer 1

Solution :-

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

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

Dividing throughout by '4'

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

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

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

Adding ( b/2 )² on both sides

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

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

[ Because a² + 2ab + b² = 2ab ]

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

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

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

Taking square root on both sides

$\implies x + \dfrac{b}{2} = \pm \sqrt{ \dfrac{ {a}^{2} }{4}}$

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

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

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

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

Hence, (a - b)/2 and (-a - b)/2 are the roots of the equation.

Answer 2

$\bigstar{\huge{\underline{\mathsf{\red{Answer}}}}}$

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

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

$\longrightarrow$$\sf{ {x}^{2} + bx = \frac{ {a}^{2} - {b}^{2} }{4}}$

$\longrightarrow$$\sf{{x}^{2} + 2( \frac{b}{2} )x + {( \frac{b}{2}) }^{2} = \frac{ {a}^{2} - {b}^{2} }{4} + { (\frac{b}{2} )}^{2} }$

$\longrightarrow$$\sf{ {(x + \frac{b}{2}) }^{2} = \frac{ {a}^{2} - {b}^{2} + {b}^{2} }{4}}$

$\longrightarrow$$\sf{ { (x + \frac{b}{2}) }^{2} = \frac{ {a}^{2} }{4} = x + \frac{b}{2} = ± \frac{a}{2} = x =± \frac{a}{2} - \frac{b}{2}}$

$\longrightarrow$$\sf{x = \frac{a - b}{2} \:or \: x = - (\frac{a + b}{2} )}$

Hence, the roots of the given equation are

$\sf{( \frac{a - b}{2} ) \: and \: - ( \frac{a + b}{2} )}$