SOLUTION: Solve the equation. Sqrt(x^2 - 24) = Sqrt(x + 6) I am lost on this one. I came up with no solution..not sure if its right.

Algebra ->  Square-cubic-other-roots -> SOLUTION: Solve the equation. Sqrt(x^2 - 24) = Sqrt(x + 6) I am lost on this one. I came up with no solution..not sure if its right.      Log On


   

Question 382368: Solve the equation.
Sqrt(x^2 - 24) = Sqrt(x + 6)
I am lost on this one.
I came up with no solution..not sure if its right.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%5E2+-+24%29+=+sqrt%28x+%2B+6%29
We need to get rid of the square roots so we can "get at" the variable. So we start by squaring both sides of the equation:
%28sqrt%28x%5E2+-+24%29%29%5E2+=+%28sqrt%28x+%2B+6%29%29%5E2
This simplifies to:
x%5E2+-+24+=+x+%2B+6
This is now a quadratic equation. So we want one side of the equation to be zero. Subtracting x and 6 from each side we get:
x%5E2-x-30+=+0
Now we factor (or use the Quadratic Formula). This factors pretty easily:
(x-6)(x+5) = 0
From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So
x-6 = 0 or x+5 = 0
Solving these we get:
x = 6 or x = -5

Whenever you solve square root equations like yours, you square both sides of the equation. This is not wrong. But doing so 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. For this reason we must check our answers.

When checking answers use the original equation:
sqrt%28x%5E2+-+24%29+=+sqrt%28x+%2B+6%29
Checking x = 6:
sqrt%28%286%29%5E2+-+24%29+=+sqrt%28%286%29+%2B+6%29
sqrt%2836+-+24%29+=+sqrt%286+%2B+6%29
sqrt%2812%29+=+sqrt%2812%29 Check!

Checking x = -5:
sqrt%28%28-5%29%5E2+-+24%29+=+sqrt%28%28-5%29+%2B+6%29
sqrt%2825-24%29+=+sqrt%28-5+%2B+6%29
sqrt%281%29+=+sqrt%281%29 Check!