Question

Solve for $x:9x^2-6ax+(a^2-b^2)=0.$

Answer 1

Given $9x^2-6ax+a^2-b^2=0$

$\Rightarrow 9x^2-6ax+(a-b)(a+b)=0$

$\Rightarrow 9x^2-3x(a-b)-3x(a+b)+(a-b)(a+b)=0$

$\Rightarrow 3x[3x-(a-b\left)\right]-(a+b)[3x-(a-b\left)\right]=0$

$\Rightarrow [3x-(a-b\left)\right][3x-(a+b\left)\right]=0$

$3x-(a-b)=0$

$3x=a-b$

$x=\frac{a-b}{3}$

$3x-(a+b)=0$

$3x=(a+b)$

$x=\frac{a+b}{3}$

$\therefore x=\frac{(a-b)}{3}$ or $x=\frac{(a+b)}{3}$