Question

`x^2-4ax-b^2+4a^2=0` 

 

Answer 1

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

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

∴`x^2-4ax-b^2+4a^2=0` 

⇒`x^2-(b+2a)x(b-2a)x-(b-2a)(b+2a)=0` 

⇒`x[x-(b+2a)]+(b-2a)[x-(b+2a)]=0` 

⇒`[x-(b+2a)] [x+(b-2a)]=0` 

⇒`x-(b+2a)=0  or  x+(b-2a)=0` 

⇒`x=2a+b  or  x=-(b-2a) `

⇒`x=2a+b  or  x=2a-b` 

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