1. ## Isolating a variable

I'm having a bit of trouble isolating "v" in the following equation:

d = (v/g) * (v + sqrt(v^2 + 2gh ) )

2. Multiply g/v to both the sides.

$\frac{dg}{v} = v + \sqrt{(v^2 + 2gh)}$

$\frac{dg}{v} - v = \sqrt{(v^2 + 2gh)}$

Square both sides and simplify.

3. This is precisely the part I need help with.

4. I am stuck at:

((dv)/g - v)^2 - v^2 = 2gh

Can I square root both sides and end up with:

(dv)/g - v - v = sqrt(2gh) ?

5. I really need help.
What do I do after ((dv)/g - v)^2 - v^2 = 2gh ?

6. $(\frac{dg}{v} - v)^2 - v^2 = 2gh$

$(\frac{dg}{v})^2 + v^2 - 2dg - v^2 = 2gh$

$(\frac{dg}{v})^2 - 2dg = 2gh$

Now solve for v.

7. Oh, thank you. Quick question though:
Isn't (dg)/v * -v = -dg rather than -2dg?

8. No. When expanding binomial squares you will get:

$(a+b)^2= a^2 + 2ab + b^2$

Remember that you get $\frac{dg}{v} \times -v = -dg$ as well as $-v \times \frac{dg}{v} = -dg$ which when you add them together you get -2dg

9. Wow, it has been a while. Thank you.

Now that I have (dg)/v = sqrt(2gh + 2dg), is it possible to slimpify but turning both into their reciprocals?

So, v/(dg) = 1/sqrt(2gh + 2dg), v = (dg)/sqrt(2gh + 2dg).

Is this correct?

10. You can further simplify it as

$v = d\sqrt{\frac{g}{2(h + d)}}$