Question

Solve by factorisation 4x^2-4ax+(a^2-b^2)

Answer 1

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

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

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


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

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

Answer 2

Answer:

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

Step-by-step explanation:

Given quadratic expression

4x²-4ax+(a²-b²)=0

=> 4x²-4ax+(a+b)(a-b)=0

/* By algebraic identity:

m²-n² =(m+n)(m-n) */

Splitting the middle term, we get

=> 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)][2x-(a-b)]=0

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

=> 2x = a+b Or 2x = a-b

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

Therefore,.

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

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