Answer
a) $x=4$ or $x=1$
b) $x=-2\pm\sqrt {13}$
Work Step by Step
(a) $x-\sqrt{x}-2=0$
Using Method I: Consider $u=\sqrt{x}, u^2=x $
$u^2-u-2=0$
$(u-2)(u+1)=0$
$\therefore u-2=0$ or $u+1=0$
$u=2$ or $u=-1$
Return to $x$: $\therefore x=4$ or $x=1$
Using Method II: Square the both sides
$x-\sqrt{x}-2=0 \rightarrow x-2=\sqrt{x}$
$\therefore x^2-4x+4=x$
$\therefore x^2-5x+4=0$
$(x-4)(x-1)=0$
$x=4$ or $x=1$
(b) $\dfrac{12}{(x-3)^2}+\dfrac{10}{(x-3)}+1=0$
Using Method I: Get a common denominator
$\dfrac{12}{(x-3)^2}+\dfrac{10(x-3)}{(x-3)^2}+\dfrac{(x-3)^2}{(x-3)^2}=0$
$\therefore 12+10(x-3)+(x-3)^2=0$
$\therefore 12+10x-30+x^2+9-6x=0$
$\therefore x^2+4x-9=0$
Use the Quadratic Rule: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\therefore x=\dfrac{-4\pm\sqrt{4^2-4(1)(-9)}}{2(1)}=\dfrac{-4\pm2\sqrt{13}}{2} =-2\pm\sqrt{13}$
Using Method II: Substitute by $u=x-3$
$\dfrac{12}{u^2}+\dfrac{10}{u}+1=0$
$\dfrac{12}{u^2}+\dfrac{10u}{u^2}+\dfrac{u^2}{u^2}=0$
$12+10u +u^2=0$
Use the Quadratic Rule: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$u=\dfrac{-10\pm\sqrt{10^2-4(1)(12)}}{2(1)}$
$=\dfrac{-10\pm\sqrt{52}}{2}=\dfrac{-10\pm2\sqrt{13}}{2} =-5\pm\sqrt{13}$
Return to $x$:
$\therefore x-3=-5\pm\sqrt{13}$
$\therefore x=-2\pm\sqrt{13}$
