# Thread: solving this equation by rearanging

1. ## solving this equation by rearanging

I am totally not understanding this rearanging business

I have this, And I have to solve for X. The reason I am having trouble is because idk what to do with all the stuff that is in the square root...

2. Originally Posted by Djuice27
I am totally not understanding this rearanging business

I have this, And I have to solve for X. The reason I am having trouble is because idk what to do with all the stuff that is in the square root...
Multiply by the whole square root

$\displaystyle \frac{6}{\sqrt{x+5}} = \sqrt{x+5}$

$\displaystyle 6 = (\sqrt{x+5})^2 = x+5$

Solve for x

3. umm...yeh...

What is the x+5 on the right doing? Is that part of the step? or the one in the middle part of the step? Yeh, im really confused.

Is this what im supposed to do next? ~_~.

$\displaystyle 1 = (\sqrt{x+5})^2 = x$

then

$\displaystyle 1 = x^2+25 = x$

then

$\displaystyle -24 = x^2 = x$

then

$\displaystyle -24 = x$

I am sure this is horrifically wrong.

4. Originally Posted by Djuice27

umm...yeh...

What is the x+5 on the right doing? Is that part of the step? or the one in the middle part of the step? Yeh, im really confused. Do you want me to say
x=19? or x =1???

or is it:

$\displaystyle 1 = (\sqrt{x+5})^2 = x$

then

$\displaystyle 1 = x^2+25 = x$

then

$\displaystyle -24 = x^2 = x$

then

$\displaystyle -24 = x$

I am sure this is horrifically wrong.
Hi Djuice27,

What e^(i*pi) is saying is this:

$\displaystyle \frac{6}{\sqrt{x+5}}=\frac{\sqrt{x+5}}{1}$

Solve the proportion.

$\displaystyle \left(\sqrt{x+5}\right)^2=6$

$\displaystyle x+5=6$

$\displaystyle x=1$

5. Mhmm, but what happens to the sqrt and power of 2?

6. Originally Posted by Djuice27
Mhmm, but what happens to the sqrt and power of 2?
Keep in mind that $\displaystyle \left(\sqrt{a}\right)^2 = a$, so $\displaystyle \left(\sqrt{x+5}\right)^2=x+5$

7. Alright, thanks. I will keep that in mind. I might not understand why, but I will remember to just make it normal next time.

8. Originally Posted by Djuice27
Alright, thanks. I will keep that in mind. I might not understand why, but I will remember to just make it normal next time.
Try a simple numerical example to test it.

$\displaystyle \left(\sqrt{2}\right)^2=\sqrt{2}\cdot \sqrt{2}=\sqrt{4}=2$

9. Ohhhh Got it. Thanks!