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Math Help - equation

  1. #1
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    Exclamation equation

    Homework due in 1 hour, please help me solve these problems....

    sqrt (x-5) + 14 = 3x + 2


    4x - x/3 + (x + 1)/9 = 2x
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tara View Post
    sqrt (x-5) + 14 = 3x + 2
    Isolate the square root:
    \sqrt{x - 5} = 3x - 12

    Now square both sides.
    x - 5 = (3x - 12)^2
    and solve for x normally from there.

    Always always always plug your solutions back into the original equation. This method occasionally gives extra solutions that don't work. If none of the solutions work, then there is no solution.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tara View Post
    Homework due in 1 hour, please help me solve these problems....

    sqrt (x-5) + 14 = 3x + 2


    4x - x/3 + (x + 1)/9 = 2x
    4x - \frac{x}{3} + \frac{x + 1}{9} = 2x

    I'm not sure what the problem is. Either add up the left hand side as fractions, or simply multiply both sides of the equation by 9:
    36x - 3x + (x + 1) = 18x

    Or did I misread the question?

    -Dan
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  4. #4
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    So, here is what I got on that problem Dan....

    x-5 + 14 = (3x+2)^2
    x-5 + 14 = (3x + 2) (3x + 2)
    x -5 + 14 = 9x^2 + 12x + 4
    set one side to zero
    0= -x + 5 - 14 + 9x^2 + 12x + 4
    0= 9x^2 + 11x -5

    and that as far as I can get... ???
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tara View Post
    So, here is what I got on that problem Dan....

    x-5 + 14 = (3x+2)^2
    x-5 + 14 = (3x + 2) (3x + 2)
    x -5 + 14 = 9x^2 + 12x + 4
    set one side to zero
    0= -x + 5 - 14 + 9x^2 + 12x + 4
    0= 9x^2 + 11x -5

    and that as far as I can get... ???
    Remember you have a square root on that x - 5. It does not cover the 14! See my first response for what I think is the best approach. If you do it your way you do not get rid of the square root:
    \left ( \sqrt{x - 5} + 14 \right ) ^2 = (x - 5) + 2 \cdot 14 \cdot \sqrt{x - 5} + 14^2

    So continuing from where I left off in my first post:
    x - 5 = (3x - 12)^2

    x - 5 = 9x^2 - 72x + 144

    9x^2 - 73x + 149 = 0

    It might factor, it might not. For something with coefficients that look like this I'd just skip right to the quadratic formula to finish it.

    (Hint: Check the discriminant. What does that tell you about the possible solutions for x?)

    -Dan
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  6. #6
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    Quote Originally Posted by tara View Post

    sqrt (x-5) + 14 = 3x + 2
    First, after rearranging, the equation equals \sqrt{x-5}=3x-12. We require x-5\ge0 & 3x-12\ge0, from here, if the solution does exist, it must lie in the interval [5,\infty[. Now note that the equation equals \sqrt{x-5}-3(x-5)=3, and for x\ge5\implies\sqrt{x-5}\le3(x-5), hence the LHS is negative, so the equation has no solution.
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  7. #7
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    thanks

    To both of you for posting. I am new on here and am just getting the "rules" down on how to do things. I appreciate your assistance...
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