how do you solve this equation (20-r)^1/2=r

- anonymous

how do you solve this equation (20-r)^1/2=r

- katieb

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- anonymous

naynay, you need to square both sides so you can get access to the r under the square root.
So\[20-r=r^2 \rightarrow r^2+r-20=0\]which is quadratic. This can now be factored as\[(r+5)(r-4)=0\]so \[r=-5\]or\[r=4\]

- anonymous

how do you solve this (6b)^1/2=(8-2b)^1/2

- anonymous

One sec.

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- anonymous

ok

- anonymous

You have to remove the square roots. You do this by squaring. Whatever you do to one side, you must do to the other, so,\[\sqrt{6b}=\sqrt{8-2b} \rightarrow 6b=8-2b \rightarrow 8b=8 \rightarrow b=1\]

- anonymous

ok so b will equal 1

- anonymous

yes

- anonymous

oh ok i see what you did

- anonymous

ok this one i really dont get -3=(37-3n)^1/2-n

- anonymous

Have you become a fan yet?

- anonymous

no i will

- anonymous

You're always looking to isolate your variable. Here I would add n to both sides and then square both sides, so\[n-3=\sqrt{37-3n} \rightarrow (n-3)^2=37-3n\]Then expand the left hand side\[n^2-6n+9=37-3n \rightarrow n^2-3n-28=0\]which you can then solve by factoring.

- anonymous

where did you get 37 from

- anonymous

You gave it to me in your question:
-3=(37-3n)^1/2-n

- anonymous

oh ok im sorry didnt look twice =)

- anonymous

np

- anonymous

thank you a lot with helping me because im really not all that good in math

- anonymous

You're welcome. All you need is practice.

- anonymous

Things will start to click.

- anonymous

yeah i have a friend of mine comin to help me twice a week so hopefully it work =)

- anonymous

I'm sure it will. Good luck, and you can use this site too for help.

- anonymous

yeah i like this site a lot it does help

- anonymous

can you help me with this (-3-4x)^1/2-(-2-2x)^1/2=1

- anonymous

\[\sqrt{-3-4x}-\sqrt{-2-2x}=1\]Okay, for ones like this, it's best to move one of the square roots to the other side and then square both sides:\[\sqrt{-3-4x}=1+\sqrt{-2-2x}\]square both sides:\[-3-4x=\left( 1+\sqrt{-2-2x} \right)^2\]Now the hard part is to expand the right-hand side\[\left( 1+\sqrt{-2-2x} \right)^2=1+2\sqrt{-2-2x}+\left( \sqrt{-2-2x} \right)^2\]\[=1+2\sqrt{-2-2x}+(-2-2x)\]\[=-1-2x+\sqrt{-2-2x}\]

- anonymous

ok so this girl helped me and we got it right =)

- anonymous

Okay, so now looking at this, we have the problem of the square root again. This is because of the expansion we had to do earlier BUT we can get rid of it using the same technique as before - move everything that has nothing to do with the square root to the other side, and then square both sides.\[-3-4x=-1-2x-\sqrt{-2-2x}\]becomes\[-2-2x=-\sqrt{-2-2x}\]which is the same as\[\sqrt{-2-2x}=2+2x \rightarrow -2-2x=(2+2x)^2\]\[-2-2x=4+8x+4x^2 \rightarrow 4x^2+10x+6=0\]i.e.\[2x^2+5x+3=0 \rightarrow (x+2)(x+3)=0 \rightarrow x=-2,-3\]

- anonymous

Well, I just thought I'd finish it.

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