Question

`4x^2-4a^2x+(a^4-b^4)=0` 

 

Answer 1

We write, `-4a^2x=-2(a^2+b^2)x-2(a^2-b^2)x  as` 

`4x^2xx(a^4-b^4)=4(a^4-b^4)x^2=[-2(a^+b^2)]x xx[-2(a^2-b^2)]x` 

∴` 4x^2-4a^2x(a^4-b^4)=0` 

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

⇒` 2x[2x-(a^2+b^2)]-(a^2-b^2)[2x-(a^2+b^2)]=0` 

⇒`[2x-(a^+b^2)][2x-(a^2-b^2)]=0` 

⇒` 2x-(a^2+b^2)=0  or  2x-(a^2-b^2)=0` 

⇒`x=(a^2+b^2)/2`  or `(a^2-b^2)/2` 

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