Question

Please help!
What is the excluded value for these radical functions???

Answer 1

Okay, by "excluded value" they apparently are referring to an extraneous solution introduced during solving by squaring both sides of the equation.

As an example (not one of your problems, but very similar)
\[\sqrt{10-2x} =x-1\]If we solve by squaring both sides and forming a quadratic equation and solving that, we get two solutions. We try both of them in the original equation, but only one will work. The other is an extraneous solution or excluded value.

\[\sqrt{10-2x} = x-1\]Square both sides\[10-2x=(x-1)^2 = x^2-2x+1\]Move all terms to one side and solve\[x^2-2x+2x+1-10 = 0\]\[x^2-9=0\]\[x=\pm3\]Now we try the solutions in the original equation:
\[x=3\]\[\sqrt{10-2(3)} = 3-1\]\[\sqrt{10-6} = 2\]\[\sqrt{4} = 2\checkmark\]That solution is valid.

\[x=-3\]
\[\sqrt{10-2(-3)} = -3-1\]\[\sqrt{10+6} = -4\]\[4 = -4\]That solution is not valid. It would be valid as a solution of \[-\sqrt{10-2x} = x-1\] (but then \(x = 3\) would not be valid).

Answer 2

In general, if you solve an equation by squaring, cubing, etc. in this fashion, you better test all of your solutions.