Solving $x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$.

Question

I have problem with this equation: $$x-\sqrt{(x^2-36)} = {(x-6)^2\over 2x+12}$$ Any ideas on beatiful solving?

Answer 1

Assume $x+6=a$ and $x-6=b$

So LHS= $(a+b)/2-\sqrt{ab}=1/2 * (\sqrt a-\sqrt b)^2$.

RHS= $b^2/2a$.

Hence, $b/\sqrt a=\sqrt a- \sqrt b$

Hence $b=a-\sqrt{ab}$

Therefore $\sqrt {ab}=a-b=12.$ Now, put $a=12+b$ and solve $b(12+b)=144$.

Hope this helps. Correct me if I'm wrong!

Answer 2

$$ \begin{array}{l l l} x-\sqrt{(x^2-36)}& =& {(x-6)^2\over 2x+12}\\ (2x+12)\left(x-\sqrt{(x^2-36)}\right)&=&(x-6)^2\\ 12x+2x^2+(-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2\\ (-12-2x)\sqrt{(x^2-36)}&=&(x-6)^2-12x-2x^2\\ (-12-2x)^2(x^2-36)&=&\left((x-6)^2-12x-2x^2\right)^2\\ 4x^4+48x^3-1728x-5184&=&x^4+48x^3+504x^2-1728x+1296\\ 3x^4-504x^2-6480&=&0\\ \text{Substitute }y=x^2\\ 3y^2-504y-6480&=&0\\ 3(y-180)(y+12)&=&0\\ (y-180)(y+12)&=&0\\ \end{array} $$ We now split into 2 equations $$y-180=0\text{ or }y+12=0$$ $$y=180\text{ or }y=-12$$ $$x^2=180\text{ or }x^2=-12$$ $\sqrt{-12}$ has no solutions $$x=\pm\sqrt{180}=\pm6\sqrt{5}$$