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Math Help - couple math 95 problems

  1. #1
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    couple math 95 problems

    [SIZE=3]& # 8 7 3 0 ;
    x-2(√(x-3) = 3
    so I substract the x from both sides but then what? do I square both sides or do I divide by 2 and make the other side a fraction?

    √(5x+1) = x+1
    I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to?

    also why does √x-3 not equal x+9 ? isn't -3*-3 = +9?

    dang is there a way to make the square root symbol extend across the numbers?
    Last edited by harnar; February 21st 2009 at 09:17 AM. Reason: didn't type what I wanted
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  2. #2
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    Quote Originally Posted by harnar View Post
    [SIZE=3]& # 8 7 3 0 ;
    x-2(√(x-3) = 3
    so I substract the x from both sides but then what? do I square both sides or do I divide by 2 and make the other side a fraction?

    √(5x+1) = x+1
    I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to?

    also why does √x-3 not equal x+9 ? isn't -3*-3 = +9?

    dang is there a way to make the square root symbol extend across the numbers?
    1. I assume that you want to solve

    x-2\sqrt{x-3}=3 If so

    -2\sqrt{x-3} =3-x~\implies~2\sqrt{x-3}=x-3

    Let x-3 = y. Then your equation becomes:

    -2y = y^2~\implies~0=y^2-2y~\implies~y(y-2)=0

    A product of 2 factors equals zero if one factor equals zero:

    y = 0~\vee~y-2=0~\implies~y=0~\vee~y=2

    That means: x-3 = 0 ==> x = 3 or x-3=2 ==> x = 5

    You must plug in the result into the original equation to check if the result is a valid solution:

    x = 3: 3-2\sqrt{3-3}=3

    x = 5: 5 - 2\sqrt{5-3} \neq 3

    Therefore: Only x = 3 is a solution.
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  3. #3
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    Quote Originally Posted by harnar View Post
    ...

    √(5x+1) = x+1
    I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to?
    ...
    1. Square both sides, you'll get:

    x^2-3x=0~\implies~x(x-3)=0

    2. A product equals zero if one factor equals zero:

    x = 0~\vee~x-3=0~\implies~\boxed{x=0~\vee~x=3}

    Plug in the results into the original equation to check if they are valid solutions:

    x = 0: \sqrt{0+1} = 0+1~\longrightarrow~OK

    x = 3: \sqrt{15+1} = 3+1~\longrightarrow~OK
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