# Homework Help: Solve for x in terms of y (Quadratic formula)

1. Sep 12, 2010

1. The problem statement, all variables and given/known data
Use the quadratic formula to solve the equation for (a) x in terms of y and (b) y in terms of x.

2. Relevant equations
4x^2 - 4xy + 1 - y^2 = 0

3. The attempt at a solution
I am not really sure where to start at all. If I could just figure out the values for a, b, and c of the quadratic formula then the rest would be simple (for me). Do you take the values from the current equation like this?

a = 4, b = -4y, c = 1 - y^2

I feel as if I am over analyzing this problem. If anyone could point me to a start that would be greatly appreciated. There are no examples from my textbook that pertain to this exact type of problem.

2. Sep 12, 2010

### Pengwuino

Yes, you got it.

3. Sep 12, 2010

Thank you for the quick reply. I did run into one more issue with this problem. After plugging in the values into the quadratic formula, I run into this step:

(4y +– sqrt(32y^2 - 16)) / 8

There are numbers inside the sqrt that I could pull out: 32 and -16. My next planned step it to re-arrange the values inside the sqrt as follows:

1) sqrt(32y^2 - 16) = sqrt((16)(2y^2) - (1)(16))
2) (16) * sqrt(2y^2 - 1)

This doesn't seem right though. I keep looking for an example, definition, rule, or law to use in this scenario but can't seem to think of one.

EDIT: Actually, it seems right now that I factored correctly.

1) sqrt(32y^2 - 16) = sqrt((16)(2y^2 - 1))
2) (16) * sqrt(2y^2 - 1)

Then the whole formula will look something like this:

(4y +- 4 * sqrt(2y^2 - 1)) / 8

I can then reduce the fraction to (y +- sqrt(2y^2 - 1)) / 2

I think I've got it! This problem has been stopping me for a while now.

4. Sep 20, 2010

### zgozvrm

Correct! Now, you've solved half the problem (you've solved for x in terms of y) ... you still need to solve for y in terms of x