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  1. #1
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    solve radical expression

    solve:
    sq rt of 5x^2 -7 = 2x
    and
    sq rt of 1-2x = 1+x
    Last edited by Belanova; March 16th 2008 at 05:12 PM. Reason: latex
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belanova View Post
    solve:
    sq rt of 5x^2 -7 = 2x
    and
    sq rt of 1-2x = 1+x
    do you mean:

    \sqrt{5x^2 - 7} = 2x

    and

    \sqrt{1 - 2x} = 1 + x?

    if so, just square both sides and continue. for example, after squaring both sides, the equation becomes 5x^2 - 7 = 4x^2. now solve for x. (be sure to check your solutions in the original equation)
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  3. #3
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    So for the first one..
    x^2 -7 =0
    +7 +7
    x=7
    no solution

    And the second one turns into
    1 - 2x = 1 + x^2
    -1 = -1
    -----------------
    -2x = x^2
    divide both by 2
    -2 = x
    no solution
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belanova View Post
    So for the first one..
    x^2 -7 =0
    +7 +7
    x=7
    no solution
    this is wrong. you have x^2 = 7, not x = 7

    And the second one turns into
    1 - 2x = 1 + x^2
    -1 = -1
    -----------------
    -2x = x^2
    divide both by 2
    -2 = x
    no solution
    you didn't write x^2 on the right the first time. make sure you are doing the right thing. should it be x^2 or x?
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  5. #5
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    the questions are how you typed them. i thought i had to square square 1 and x..
    1-2x = 1 + 2x + x^2
    ..
    -4x = x^2 divide by x
    -4 = x ; no solution
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belanova View Post
    the questions are how you typed them. i thought i had to square square 1 and x..
    1-2x = 1 + 2x + x^2
    ..
    -4x = x^2 divide by x
    -4 = x ; no solution
    what is the solution for the first?

    you cannot divide by x like that.


    1 - 2x = 1 + 2x + x^2

    \Rightarrow x^2 + 4x = 0

    \Rightarrow x(x + 4) = 0

    \Rightarrow x = 0 or x = -4

    (since x = 0, it was invalid for you to divide by it. you need to factor by pulling out the common term).

    now continue
    Last edited by Jhevon; March 16th 2008 at 07:39 PM.
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  7. #7
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    no solution for both? :|

    1. x^2 =7 cnt divide by x
    2. x+4=0
    x=-4 = no solution
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Belanova View Post
    no solution for both? :|

    1. x^2 =7 cnt divide by x
    2. x+4=0
    x=-4 = no solution
    no. if you have x^2 on one side of an equation, you can square root both sides, as long as the other side is non-negative.

    x^2 = 7 \implies x = \pm \sqrt{7}. check both these solutions.

    you have x = 0 and x = -4 as possible solutions to the second. check both these solutions
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