SOLUTION: I have been trying to solve for x in the following equation: 10 / sq.root x-5 = sq.root x-5 -3 Not really sure how to better display the square roots with text, but in the

Algebra ->  Radicals -> SOLUTION: I have been trying to solve for x in the following equation: 10 / sq.root x-5 = sq.root x-5 -3 Not really sure how to better display the square roots with text, but in the      Log On


   

Question 241647: I have been trying to solve for x in the following equation:
10 / sq.root x-5 = sq.root x-5 -3
Not really sure how to better display the square roots with text, but in the second number, after the equals sign, the x-5 is in the root, and the -3 is not.
Thanks for your help!
Pat
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
10+%2Fsqrt%28+x-5%29++=++sqrt%28+x-5+%29+-3
Both the fraction and the square roots make solving this equation more complicated. So we will start by working toward eliminating both.

A lot can be accomplished by multiplying both sides by sqrt%28x-5%29:
sqrt%28x-5%29%2810+%2Fsqrt%28+x-5%29%29++=++sqrt%28x-5%29%28sqrt%28+x-5+%29+-3%29
On the right side we need to use the Distributive Property:

On the left side the square roots cancel each other. On the right side, let's see what happens:
10++=++%28sqrt%28x-5%29%29%5E2+-+3sqrt%28x-5%29
10+=+x+-+5+-+3sqrt%28x-5%29
By multiplying by sqrt%28x-5%29 we have eliminated the fraction and one of the two square roots! Now we need to eliminate the remaining square root. To do this we need to isolate the square root. So we will subtract x and add 5:
-x+%2B+15+=+-3sqrt%28x-5%29
Next we square bot sides:
%28-x+%2B+15%29%5E2+=+%28-3sqrt%28x-5%29%29%5E2
x%5E2+-30x+%2B+225+=+9%28x-5%29
x%5E2+-+30x+%2B+225+=+9x+-+45
And the last square root is gone. We now have a quadratic equation to solve. So we will get one side equal to zero by subtracting 9x and adding 45:
x%5E2+-39x+%2B+270+=+0

(I'm sorry to say I do not not time right now to show the rest of the solution. I will describe the steps you should take. If I get time later I will come back and finish this.)
To finish this:
  1. The equation x%5E2+-39x+%2B+270+=+0 will not factor so you will need to use the Quadratic Formula.
  2. The Quadratic Formula should give you two solutions.
  3. For two reasons it becomes important, not just a good idea, to check your answers. These two reasons are:
    • With square roots in the original equation we must make sure that each solution does not make the number inside the square root (aka the radicand) negative. If a solution makes a radicand negative then we must reject it.
    • We squared both sides of the equation as we solved it. Squaring both sides can introduce extraneous solutions. (Exptraneous solutions are solutions that do not actually fit the original equation.) So we have to check our answers to make sure they actually work.
    Either of these by itself makes it important to check our answers. We have both in this problem. When checking your answers, be sure to use the original equation.