Answer:
(a) x = 12
Step-by-step explanation:
The radicals in this equation can be eliminated simply by squaring both sides. Then the resulting linear equation is solved in two steps.
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square
[tex]\left(\sqrt{2x-8}\right)^2=\left(\sqrt{x+4}\right)^2\\\\2x-8=x+4\qquad\text{simplify}\\\\x-8=4\qquad\text{subtract $x$}\\\\\boxed{x=12}\qquad\text{add 8}[/tex]
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Additional comment
You know the contents of each radical must not be negative. That means 2x-8 ≥ 0, or x≥4. We know x=4 doesn't work in the equation (0≠√8), so we only need to check the answers x=12 and x=7. Of those, only x=12 satisfies the equation.
