Question

${\displaystyle x^2-4ax-b^2+4a^2=0}$ 

 

Answer 1

We write, ${\displaystyle -4ax=-(b+2a)x+(b-2a)xas}$ 

${\displaystyle x^2\times (-b^2+4a^2)=(-b^2+4a^2)x^2=-(b+2a)x\times (b-2a)x}$ 

∴${\displaystyle x^2-4ax-b^2+4a^2=0}$ 

⇒${\displaystyle x^2-(b+2a)x(b-2a)x-(b-2a)(b+2a)=0}$ 

⇒${\displaystyle x[x-(b+2a)]+(b-2a)[x-(b+2a)]=0}$ 

⇒${\displaystyle [x-(b+2a)][x+(b-2a)]=0}$ 

⇒${\displaystyle x-(b+2a)=0 \text{or} x+(b-2a)=0}$ 

⇒${\displaystyle x=2a+b \text{or} x=-(b-2a)}$

⇒${\displaystyle x=2a+b \text{or} x=2a-b}$ 

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