Question

Solve by factorizaton: 4x² - 4a²x + (a^4 - b^4) =0

Answer 1

Given:

4 x^2 - 4 a^2 x + ( a^4 - b^4 )

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

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

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

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

==> Either

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

==> 2 x = a^2 + b^2

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

or ,

==> 2 x - a^2 + b^2 =0

==> 2 x = a^2 - b^2

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

The solutions are either x = (a^2 + b^2 ) /2 or x = (a^2 - b^2)/2

Hope it helps you

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