1. #1
    Banned sandmoth12's Avatar
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    Test your algebra skills!

    I realize it's against the rules to post homework so I will post the question and the answer as I have already did all the work. You just have to show the work.

    square root(2y-3)...subtract...square root(3y+3)...add...square root(3y-2)=0

    The answer is y=2.

    EDIT: This one was a little tough. Took a few tries to get it right. It's definitely one of the longer problems so attention to detail when playing with signs helps.

    ---------- Post added 2012-01-29 at 03:55 PM ----------

    No one? Am I the only lonely math nerd on this site =(
    Last edited by sandmoth12; 2012-01-29 at 11:29 PM.

  2. #2
    You should end up with 2 possible answers to this one (unless something about the problem rules out a negative solution). I will post my solution briefly, but I want to give you a chance to work on that first.

    Edit: Oh, of course, the starting equation would be non-real if y is negative. Silly me.

    Here's my solution, then:
    Hm... let's see. This might not be the most elegant way to solve the problem, but here's how I'd solve it:
    sqrt(2y-3)-sqrt(3y+3)+sqrt(3y-2)=0
    Add sqrt(3y+3) to both sides:
    sqrt(2y-3)+sqrt(3y-2)=sqrt(3y+3)
    Square both sides:
    2y-3+2sqrt((2y-3)(3y-2))+3y-2=3y+3
    Combine like terms:
    5y-5+2sqrt(6y^2-13y+6)=3y+3
    Subtract 5y and add 5 to both sides:
    2sqrt(6y^2-13y+6)=-2y+8
    Divide by 2:
    sqrt(6y^2-13y+6)=-y+4
    Square both sides:
    6y^2-13y+6=y^2-8y+16
    Subtract the right side:
    5y^2-5y-10=0
    You could solve this trinomial directly without too much trouble, but I'm going to divide by 5 first to make it look a little nicer:
    y^2-y-2=0
    Factors into:
    (y-2)(y+1)=0
    So y=2 or y=-1. If you're not dealing with imaginary/complex numbers, we can throw out the y=-1 because that would have given us some square roots of negatives in the first equation. So yes, y=2.
    Last edited by Meteoric; 2012-01-30 at 12:03 AM.

  3. #3
    So....

    Root(2y-3)-Root(3y+3)+Root(3y-2) = 0
    Square both sides
    (2y-3)-(3y+3)+(3y-2) = 0
    2y-3-3y-3+3y-2=0
    combine terms
    2y-8=0
    2y=8
    y=4

    What did I do wrong?

  4. #4
    Quote Originally Posted by Totle View Post
    What did I do wrong?
    (a+b+c)^2=/=a^2+b^2+c^2
    You can't just distribute the squared to each term when there's addition/subtraction in the way, you'd have to multiply them all out and you'd get mixed terms (like FOIL, but with 3 things in each parenthesis instead of 2). That's why I moved one of the square roots to the other side first - to avoid getting so many mixed terms when squaring both sides.
    Last edited by Meteoric; 2012-01-30 at 12:11 AM.

  5. #5
    Banned sandmoth12's Avatar
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    Quote Originally Posted by Meteoric View Post
    You should end up with 2 possible answers to this one (unless something about the problem rules out a negative solution). I will post my solution briefly, but I want to give you a chance to work on that first.

    Edit: Oh, of course, the starting equation would be non-real if y is negative. Silly me.

    Here's my solution, then:
    Hm... let's see. This might not be the most elegant way to solve the problem, but here's how I'd solve it:
    sqrt(2y-3)-sqrt(3y+3)+sqrt(3y-2)=0
    Add sqrt(3y+3) to both sides:
    sqrt(2y-3)+sqrt(3y-2)=sqrt(3y+3)
    Square both sides:
    2y-3+2sqrt((2y-3)(3y-2))+3y-2=3y+3
    Combine like terms:
    5y-5+2sqrt(6y^2-13y+6)=3y+3
    Subtract 5y and add 5 to both sides:
    2sqrt(6y^2-13y+6)=-2y+8
    Divide by 2:
    sqrt(6y^2-13y+6)=-y+4
    Square both sides:
    6y^2-13y+6=y^2-8y+16
    Subtract the right side:
    5y^2-5y-10=0
    You could solve this trinomial directly without too much trouble, but I'm going to divide by 5 first to make it look a little nicer:
    y^2-y-2=0
    Factors into:
    (y-2)(y+1)=0
    So y=2 or y=-1. If you're not dealing with imaginary/complex numbers, we can throw out the y=-1 because that would have given us some square roots of negatives in the first equation. So yes, y=2.
    Awesome man nearly the exact same way I did it. I don't have any candy but instead /salute =)
    I plugged -1 back into the equation and it didn't checkout so I just dropped it.

    ---------- Post added 2012-01-29 at 04:10 PM ----------

    Quote Originally Posted by Totle View Post
    So....

    Root(2y-3)-Root(3y+3)+Root(3y-2) = 0
    Square both sides
    (2y-3)-(3y+3)+(3y-2) = 0
    2y-3-3y-3+3y-2=0
    combine terms
    2y-8=0
    2y=8
    y=4

    What did I do wrong?
    Looks like you didn't isolate a single square root on one side of the equal sign before you squared everything.
    Last edited by sandmoth12; 2012-01-30 at 12:16 AM.

  6. #6
    Herald of the Titans kailtas's Avatar
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    I always get sad when i see math on these forums... we use different words in Norway -_-
    Your greed, your foolishness has brought you to this end.

    - Prince Malchezaar

  7. #7
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    Yeah, solved it pretty much same way as Meteoric did. Took a few tries to reach the correct answer, and even had to look up some of the rules like:

    sqrt(a) x sqrt(b) = sqrt(ab)

    Been a long time since I solved these kind of problems.

  8. #8

  9. #9
    Quote Originally Posted by kailtas View Post
    I always get sad when i see math on these forums... we use different words in Norway -_-
    So much for maths being the universal language :P

  10. #10
    Oh balls, it's been over a decade since algebra. Hats off to you guys who got it.

  11. #11
    enter this in solve function on my casio calc. +10 internnets?
    FT for life <3

  12. #12
    High Overlord Rhidios's Avatar
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    I promise I didn't look at the answers above me

    sqrt(2y-3) - sqrt(3y+3) + sqrt(3y-2) = 0

    sqrt(2y-3) + sqrt(3y-2) = sqrt(3y+3)

    2y-3 + 2*sqrt(6y² -13y +6) + 3y-2 = 3y+3

    2*sqrt(6y² -13y +6) = -2y + 8

    sqrt(6y² -13y +6) = -y + 4

    6y² - 13y + 6 = y² - 8y + 16

    5y² - 5y - 10 = 0

    y² - y - 2 = 0

    (y - 2)(y + 1) = 0

    y = 2 V y = -1

    No real solution for y = -1, so y = 2


    I messed up the first time because I made a typo and typed a + instead of a -...
    Last edited by Rhidios; 2012-01-30 at 02:38 PM.

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