SOLUTION: how do you solve the cubic root of (little 3) 3, above the square root of 2x^2+6x-6=(little 3 again), 3, above the square root of x^2+1

Question 615072: how do you solve the cubic root of (little 3) 3, above the square root of 2x^2+6x-6=(little 3 again), 3, above the square root of x^2+1
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

This equation is, in essence. a proportion (i.e. a fraction = a fraction). So we can start by cross-multiplying:

We can get rid of the cube roots by simply dividing both sides by :

We can get rid of the square roots by squaring both sides:

This is a quadratic equation. So we want one side to be a zero. Subtracting and 1 from each side we get:

Now we factor (or use the Quadratic Formula). This factors easily:

From the Zero Product Property we know that one or more factors must be zero:
or
Solving these we get:
or

We must check our solutions. It is not optional! At one point we squared both sides of the equation. This is not wrong. But it can introduce what are called extraneous solutions. Extraneous solutions are solutions that work in the squared equation but do not work in the original equation. The only way to find extraneous solutions is to check. (If we find any we reject/discard them.)

Use the original equation to check:

Checking x = 5:

Simplifying:

Check fails! x = 5 must be an extraneous solution. So we discard it.

Checking x = 1:

Simplifying:

Check!

So the only solution to your equation is x = 1.
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