Question

Given \(\displaystyle \frac{\left(\sqrt{x+4}+\sqrt{x-10}\right)^2}{(x+4)-(x-10)}=\frac{5}{2}\), find x.

Hint:

After expanding \((\sqrt{u}+\sqrt{v})^2\), isolate the radical before squaring. Always check the domain to discard any extraneous root.

Answer 1

\[ \frac{(\sqrt{x+4}+\sqrt{x-10})^2}{(x+4)-(x-10)} =\frac{(x+4)+(x-10)+2\sqrt{(x+4)(x-10)}}{14} =\frac{2x-6+2\sqrt{(x+4)(x-10)}}{14}. \] Set this equal to \(\frac{5}{2}\) and clear denominators: \[ 2x-6+2\sqrt{(x+4)(x-10)}=35 \quad\Longrightarrow\quad \sqrt{(x+4)(x-10)}=\frac{41-2x}{2}. \] Square both sides (noting \(x\ge 10\)): \[ (x+4)(x-10)=\frac{(41-2x)^2}{4} \ \Longrightarrow\ x^2-6x-40=x^2-41x+\frac{1681}{4}. \] Hence \(35x=\frac{1841}{4}\), so \[ x=\frac{1841}{140}=\frac{263}{20}. \] Domain check: \(10\le x\le 20.5\) from the radical equation; \(x=\frac{263}{20}=13.15\) is valid. \[ \boxed{\dfrac{263}{20}} \]