SOLUTION: I just can't figure out the first step to get this set up for a quadratic equation. (x / 5 - 3)^2 = 63

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Question 9832: I just can't figure out the first step to get this set up for a quadratic equation.
(x / 5 - 3)^2 = 63
Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
The first step is to "undo" the square by taking the square root of both sides. That is, unless you want to multiply out the quantity squared, and make a REALLY hard problem out of it. Remember, any time you take the square root of both sides, you MUST include a PLUS or MINUS sign, or you will be losing half of the solutions everytime you do this.

As an example to warm up, suppose you take x%5E2+=+9. You KNOW there are two solutions for this quadratic equation. In your head you can see that the solutions are x= 3 or x = -3. If you take the square root of both sides of the equation x%5E2+=+9, you will need to include x+=+0+%2B-+3.

Now, here is the problem at hand. Begin by taking the square root of each side:
%28x%2F5-3%29%5E2+=+63
%28x%2F5-3%29+=+0+%2B-+sqrt%2863%29+

+x%2F5+-+3+=+0+%2B-+sqrt%289%29+%2A+sqrt%287%29+
+x%2F5+-+3+=+0+%2B-+3+%2A+sqrt%287%29

Next, add +3 to each side of the equation:
+x%2F5+-3+%2B+3+=+3+%2B-+3%2Asqrt%287%29
+x%2F5+=+3+%2B-+3%2A+sqrt%287%29+

Multiply both sides by 5:
x+=+5%283+%2B-+3%2Asqrt%287%29%29+

This is an acceptable answer, but probably not the one you are looking for. You probably should either multiply it out or factor it completely:
+x+=+15+%2B-+15%2Asqrt%287%29 or x+=+15%281+%2B-sqrt%287%29%29+

R^2 at SCC