# Thread: Help with this formula

1. ## Help with this formula

Hello all,

I am new to this forum so I hope I am posting correctly. Its been years since I have done any "real" math, I use math every day but its been just basic formulas and nothing too advanced. I have this formula below that solves for X but I need to solve for Y. I thought this would be simple but for some reason can't seem to crack this nut. Many of my friends have been unable as well. Any suggestions on how to do this?

X=a(d/4y – 1/√2y)

Thanks

BC

2. A few comments.

First of all, this thread belongs in Pre-University Math Help/Pre-Algebra and Algebra subforum, not Linear and Abstract Algebra.

Second of all, is this the equation you're trying to solve:

$X=a\left(\frac{d}{4y}-\frac{1}{\sqrt{2}\,y}\right)$? Or is it this:

$X=a\left(\frac{d}{4y}-\frac{1}{\sqrt{2y}}\right)$?

I'm asking what's inside the square root. That will drastically affect your answer.

3. Good question, I can see why it was confusing. Its the latter equation, the y is under the square root.

Thanks

BC

4. All right. First thing I would do is solve for the square root. What do you get when you do that?

5. Honestly that is where I am hung up. I can't seem to get rid of the square root.

6. That is why I recommended that you solve for it. If you solve for it, then you can square both sides of the equation to get rid of it. Make sense? I get the following:

$\displaystyle{X-\frac{ad}{4y}=-\frac{a}{\sqrt{2y}}.}$

Knowing I want to square next, I'd probably get a common denominator on the LHS, and change the signs:

$\displaystyle{\frac{ad-4yX}{4y}=\frac{a}{\sqrt{2y}}.}$

Can you continue?

7. Am I good so far?

a²d² - 2ad4yx + 16y²x² = a²8y

8. You could simplify it a bit, but yes, I buy that.

9. Originally Posted by troutbwc
Am I good so far?

a²d² - 2ad4yx + 16y²x² = a²8y
And now think of that as a quadratic in y and solve using the quadratic formula.