Question

Solve the equation. Identify any extraneous solutions.

x=sqrt2x+24


6 and 4 are both extraneous solutions.

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

6 and –4 are solutions.

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

Answer 1

`x=√((2x+24))\\\\The\ domain:\\2x+24\geq0\to x\geq-12\\x\geq0\\D:x\geq0\\\\x=√(2x+24)\ \ \ \ |square\ both\ sides\\\\x^2=(√(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\\otin D\ \vee\ x=6\in D`

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

Answer 2

`x=√(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=√(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\\ot \geq 0`

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