Question

Factorise :-

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

Answer 1

Step-by-step explanation:

ANSWER

The given quadratic equation is 4x

2

−4ax+(a

2

−b

2

)=0

4x

2

−4ax+(a+b)(a−b)=0

⇒4x

2

+[−2a−2a+2b−2b]x+(a−b)(a+b)=0

⇒4x

2

+(2b−2a)x−(2a+2b)x+(a−b)(a+b)=0

⇒4x

2

+2(b−a)x−2(a+b)x+(a−b)(a+b)=0

⇒2x[2x−(a−b)]−(a+b)[2x−(a−b)]=0

⇒[2x−(a−b)][2x−(a+b)]=0

⇒2x−(a−b)=0,2x−(a+b)=0

⇒x=

2

a+b

,

2

a−b

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.

Answer 2

Answer:

$\huge\bold\red{Question:-}$

4x²+4bx-(a²-b²) = 0

$\huge\bold\blue{\underline{\underline{{Solution:-}}}}$

=> 4x²+4bx-(a+b)(a-b)=0

=> 4x² +2{(a+b) - (a-b)} x -(a+b) (a-b)=0

=> 4x² +2(a+b)x -2(a-b)x -(a+b) (a-b) = 0

=> 2x {2x +(a+b)} {(a-b) {2x +(a+b)} = 0

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

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

answer.