# When does this derivative equal 7

1. Oct 4, 2009

### fghtffyrdmns

1. The problem statement, all variables and given/known data

Solving a derivative for when it equals to 7.

2. Relevant equations

$$p'(t) = {88t+210}/{2\sqrt{44t^{2}+210t}$$

3. The attempt at a solution

$$7(2\sqrt{44t^{2}+210t}) = {88t+210}$$

This is where I get stuck. Do I divied both sides by 7 then again by 2? then I would square both sides to get (88/14 - 15)^{2}. Then I expand, set it to 0 and use quadratic formula?

2. Oct 4, 2009

### Mentallic

I'm a bit confused.
At first glance I would assume this means "find the derivative where the variable t=7" but from what you've done, I guess it must instead mean "find the value(s) of t where the derivative is 7". Is this correct?

Now, without using parenthesis it's impossible to be certain what the derivative function actually is:
but judging by your last line:
I'm now assuming you mean $$p'(t) = \frac{88t+210}{2\sqrt{44t^{2}+210t}}$$
Again, is this correct?

Finally, if all is correct thus far, yes that is exactly how you would go about solving it But remember to check your answers because once you square both sides, you might have extra solutions that don't work.

3. Oct 4, 2009

### fghtffyrdmns

Yes! that is what i meant :).

The problem is, though, when I get to this:

$$7(2\sqrt{44t^{2}+210t}) = {88t+210}$$

I divide out the seven on both side to get
$$2\sqrt{44t^{2}+210t} = {88/7t+30}$$

then again by 2

$$\sqrt{44t^{2}+210t} = {88/14t+15}$$

Now, when I square both side? do I write it like this:$${44t^{2}+210t = ({88/14t+15})^{2}$$

which I just expand out?

Last edited: Oct 4, 2009
4. Oct 4, 2009

### Bohrok

You can expand it. Don't forget that you can reduce 88/14.

5. Oct 4, 2009

### fghtffyrdmns

How come I can't just take the square of 44t/14 and 15?

6. Oct 4, 2009