When does this derivative equal 7

[fghtffyrdmns]

Homework Statement

Solving a derivative for when it equals to 7.

Homework Equations

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

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?

 

[Mentallic]

I'm a bit confused.

Solving a derivative for when it equals to 7.

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:

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

but judging by your last line:

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

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.

 

[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?

 

[Bohrok]

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

 

[fghtffyrdmns]

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

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

 

[sara_87]

because (a+b)^2 does not equal a^2+b^2
in fact it is: a^2+2ab+b^2
try with numbers. (3+4)^2 doesn't equal 3^2+4^2