Solve for x in terms of y (Quadratic formula)

adillhoff

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.

Pengwuino

Yes, you got it.

adillhoff

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.

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