Question

Solve the equation. Identify any extraneous solutions.

Answer 1

$x=\sqrt{2x+24}\\ x\geq 0 \wedge 2x+24\geq0\\ x\geq 0 \wedge 2x\geq-24\\ x\geq 0 \wedge x\geq-12\\ x\geq0\\\\ x=\sqrt{2x+24}\\ x^2=2x+24\\ x^2-2x-24=0\\ x^2+4x-6x-24=0\\ x(x+4)-6(x+4)=0\\ (x-6)(x+4)=0\\ x=6 \vee x=-4\\\\ -4\not \geq 0$

So, 6 is a solution to the original equation. The value –4 is an extraneous solution.

Answer 2

$x=\sqrt{(2x+24)}\\\\The\ domain:\\2x+24\geq0\to x\geq-12\\x\geq0\\D:x\geq0\\\\x=\sqrt{2x+24}\ \ \ \ |square\ both\ sides\\\\x^2=(\sqrt{2x+24)})^2\\\\x^2=2x+24\ \ \ |-2x-24\\\\x^2-2x-24=0\\\\x^2+4x-6x-24=0\\\\x(x+4)-6(x+4)=0\\\\(x+4)(x-6)=0\iff x+4=0\ \vee\ x-6=0\\\\x=-4\notin D\ \vee\ x=6\in D$

Answer: 6 is a solution to the original equation. The value –4 is an extraneous solution.