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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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\[r=4\]

  2. anonymous
    • 5 years ago
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    how do you solve this (6b)^1/2=(8-2b)^1/2

  3. anonymous
    • 5 years ago
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    One sec.

  4. anonymous
    • 5 years ago
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    ok

  5. anonymous
    • 5 years ago
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    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,\[\sqrt{6b}=\sqrt{8-2b} \rightarrow 6b=8-2b \rightarrow 8b=8 \rightarrow b=1\]

  6. anonymous
    • 5 years ago
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    ok so b will equal 1

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    oh ok i see what you did

  9. anonymous
    • 5 years ago
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    ok this one i really dont get -3=(37-3n)^1/2-n

  10. anonymous
    • 5 years ago
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    Have you become a fan yet?

  11. anonymous
    • 5 years ago
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    no i will

  12. anonymous
    • 5 years ago
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    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.

  13. anonymous
    • 5 years ago
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    where did you get 37 from

  14. anonymous
    • 5 years ago
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    You gave it to me in your question: -3=(37-3n)^1/2-n

  15. anonymous
    • 5 years ago
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    oh ok im sorry didnt look twice =)

  16. anonymous
    • 5 years ago
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    np

  17. anonymous
    • 5 years ago
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    thank you a lot with helping me because im really not all that good in math

  18. anonymous
    • 5 years ago
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    You're welcome. All you need is practice.

  19. anonymous
    • 5 years ago
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    Things will start to click.

  20. anonymous
    • 5 years ago
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    yeah i have a friend of mine comin to help me twice a week so hopefully it work =)

  21. anonymous
    • 5 years ago
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    I'm sure it will. Good luck, and you can use this site too for help.

  22. anonymous
    • 5 years ago
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    yeah i like this site a lot it does help

  23. anonymous
    • 5 years ago
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    can you help me with this (-3-4x)^1/2-(-2-2x)^1/2=1

  24. anonymous
    • 5 years ago
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    \[\sqrt{-3-4x}-\sqrt{-2-2x}=1\]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\[\left( 1+\sqrt{-2-2x} \right)^2=1+2\sqrt{-2-2x}+\left( \sqrt{-2-2x} \right)^2\]\[=1+2\sqrt{-2-2x}+(-2-2x)\]\[=-1-2x+\sqrt{-2-2x}\]

  25. anonymous
    • 5 years ago
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    ok so this girl helped me and we got it right =)

  26. anonymous
    • 5 years ago
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    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.\[2x^2+5x+3=0 \rightarrow (x+2)(x+3)=0 \rightarrow x=-2,-3\]

  27. anonymous
    • 5 years ago
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    Well, I just thought I'd finish it.

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