anonymous
  • anonymous
how do you solve this equation (20-r)^1/2=r
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
how do you solve this (6b)^1/2=(8-2b)^1/2
anonymous
  • anonymous
One sec.

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anonymous
  • anonymous
ok
anonymous
  • 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
  • anonymous
ok so b will equal 1
anonymous
  • anonymous
yes
anonymous
  • anonymous
oh ok i see what you did
anonymous
  • anonymous
ok this one i really dont get -3=(37-3n)^1/2-n
anonymous
  • anonymous
Have you become a fan yet?
anonymous
  • anonymous
no i will
anonymous
  • 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
  • anonymous
where did you get 37 from
anonymous
  • anonymous
You gave it to me in your question: -3=(37-3n)^1/2-n
anonymous
  • anonymous
oh ok im sorry didnt look twice =)
anonymous
  • anonymous
np
anonymous
  • anonymous
thank you a lot with helping me because im really not all that good in math
anonymous
  • anonymous
You're welcome. All you need is practice.
anonymous
  • anonymous
Things will start to click.
anonymous
  • anonymous
yeah i have a friend of mine comin to help me twice a week so hopefully it work =)
anonymous
  • anonymous
I'm sure it will. Good luck, and you can use this site too for help.
anonymous
  • anonymous
yeah i like this site a lot it does help
anonymous
  • anonymous
can you help me with this (-3-4x)^1/2-(-2-2x)^1/2=1
anonymous
  • 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
  • anonymous
ok so this girl helped me and we got it right =)
anonymous
  • 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
  • anonymous
Well, I just thought I'd finish it.

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