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Math Help - Solving for X

  1. #1
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    Solving for X

    Here is the problem I have: -7 x [sqr root of (3x-8)] + 3 = -4x + 7. The answer I am looking for is 8. I first subtracted 3 then divided the right side by -7. Then I squared both sides and attempted to simplify. I didn't come up with the right answer.
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Solving for X

    -7\sqrt{3x-8} = -4x+4

    \sqrt{3x-8} = -\dfrac{(-4x+4)}{7} = \dfrac{4}{7}(x-1)

    3x-8 = \left(\dfrac{4}{7}\right)^2 \cdot (x-1)^2

    Do you follow up to there? If so, what do you get on the RHS?
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Solving for X

    Note:
    Be sure you assure that your solution x\geq\frac{8}{3}
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    Re: Solving for X

    Yes, I follow up to there. I personally didn't break up the denominator into a fraction as you did. I get 16/49 times x^2 - 2x + 1 on the RHS. I tried distributing the fraction over and I got the wrong answer again... I don't know what I'm doing wrongly. I've worked this at least 10 times, coming with the quadratic and plugging 8 in, and haven't been successful.
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  5. #5
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    Re: Solving for X

    Well, if you're having problems with fractions, then just square both sides:
    [-7SQRT(3x - 8)] squared = 49(3x - 8)
    [-4x + 4] squared = 16x^2 + 16x + 16
    So:
    147x - 392 = 16x^2 + 16x + 16
    Solve using quadratic formula.
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    MHF Contributor Siron's Avatar
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    Re: Solving for X

    Quote Originally Posted by benny92000 View Post
    I personally didn't break up the denominator into a fraction as you did. I get 16/49 times x^2 - 2x + 1 on the RHS.
    This is correct for the RHS, multiply both sides with 49 so you're 'losing' the denominator in the RHS:
    49(3x-8)=16(x^2-2x+1)

    @Wilmer:
    (-4x+4)^2=16x^2-32x+16
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  7. #7
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    Re: Solving for X

    I came up with the right answer. I had a binomial and I was squaring each term individually rather than the binomial as a whole... -.- Thanks guys.
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  8. #8
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    Re: Solving for X

    Hello, benny92000!

    -7\sqrt{3x-8}+ 3 \:=\: -4x + 7

    Whenever possible, I would avoid fractions.

    Subtract 3 from both sides: . -7\sqrt{3x-8} \:=\:-4x + 4

    Square both sides: . \bigg[-7\sqrt{3x-8}\bigg]^2 \:=\:\bigg[-4x + 4\bigg]^2

    . . . . . . . . . . . . . . . . . . 49(3x-8) \:=\:16x^2 - 32x + 16

    Simplify:. a . . . . 16x^2 - 179x + 408 \:=\:0

    Factor:. - . . . . . . (x-8)(16x-51) \:=\:0


    And we have: . \begin{Bmatrix}x - 8 \:=\: 0 & \Rightarrow & x \:=\: 8 \\ 16x - 51 \:=\: 0 & \Rightarrow & x \:=\: \frac{51}{16} \end{Bmatrix}

    . . and both answers check out!

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  9. #9
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    Re: Solving for X

    sorry thought i saw something but now i can't delete this post.could anyone do it?
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  10. #10
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    Re: Solving for X

    Posts can NOT be deleted after being answered.
    They may be useful to someone else who is learning...get it?
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