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Math Help - Solving Radical expressions

  1. #1
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    Solving Radical expressions

    I think that I did this one correctly as well but I could not figure out the factoring.
    √x+2 + √3x+7=1
    √3x+7 = 1-√x+2
    √(3x+7)^2 =(1-√x+2)^2
    3x+7=1-x = 2-2√x+2
    3x+7 = -x+3-2√x+2
    2x +4 = (-2√x+2)
    (2x+4)^2= (2√x+2)^2
    4x^2+16x+16=4(x+2)
    4x^2+12x-8=0
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  2. #2
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    Quote Originally Posted by roachbeats View Post
    I think that I did this one correctly as well but I could not figure out the factoring.
    √x+2 + √3x+7=1
    √3x+7 = 1-√x+2
    √(3x+7)^2 =(1-√x+2)^2
    3x+7=1-x = 2-2√x+2
    ...
    \left(\sqrt{3x+7}\right)^2=\left(1-\sqrt{x+2}\right)^2

    3x+7=1-2\sqrt{x+2} + x+2

    \left(2x+4\right)^2 = \left(\sqrt{x+2}\right)^2

    4x^2+16x+16 = x+2

    4x^2+15x+14=0

    x=\dfrac{-15\pm\sqrt{(-15)^2-4\cdot 4 \cdot 14}}{2\cdot 4}

    x=-2~\vee~x = -\frac74

    Now plug in these solutions into the original equation and check which is a correct solution:

    x = -2~\implies~\sqrt{-2+2} + \sqrt{3\cdot(-2)+7} = 1 Therefore \boxed{x = -2} is a solution.

    x = -\frac74~\implies~\sqrt{-\frac74+2} + \sqrt{3\cdot \left(-\frac74 \right)+7} = \frac12 + \sqrt{\frac74}\neq 1 Therefore x = -\frac74 is no solution.
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