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Math Help - equation w. radicals and exponenets

  1. #1
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    equation w. radicals and exponenets

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

    \left(\sqrt{2x+2}\right)^{2}=\left(2+\sqrt{x-3}\right)^{2}

    x+1=4\sqrt{x-3}

    i dont know where to go from here, i tried dividing by 4 and mult. the left by itself, and get lost, the answer is 7
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    Re: equation w. radicals and exponenets

    Quote Originally Posted by mathmathmathmathmathmathm View Post
    \sqrt{2x+2}-\sqrt{x-3}=2

    \left(\sqrt{2x+2}\right)^{2}=\left(2+\sqrt{x-3}\right)^{2}

    x+1=4\sqrt{x-3}

    i dont know where to go from here, i tried dividing by 4 and mult. the left by itself, and get lost, the answer is 7
    \displaystyle \begin{align*} \sqrt{2x+2} - \sqrt{x-3} &= 2 \\ \sqrt{2x+2} &= 2+\sqrt{x-3} \\ \left(\sqrt{2x+2}\right)^2 &= \left(2+\sqrt{x-3}\right)^2 \\ 2x+2 &= 4 + 4\sqrt{x-3} + x - 3 \\ x + 1 &= 4\sqrt{x-3} \\ (x + 1)^2 &= \left(4\sqrt{x-3}\right)^2 \\ x^2 + 2x + 1 &= 16(x - 3) \\ x^2 + 2x + 1 &= 16x - 48 \\ x^2 - 14x + 49 &= 0 \\ (x - 7)^2 &= 0 \\ x - 7 &= 0 \\ x &= 7  \end{align*}
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  3. #3
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    Re: equation w. radicals and exponenets

    You should of course, check the solution. (Squaring both sides of an equation may introduce "extraneous" solutions.)

    \sqrt{2(7)+ 2}= \sqrt{14+ 2}= \sqrt{16}= 4
    \sqrt{7- 3}= \sqrt{4}= 2

    4- 2= 2 so the solution checks.
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    Re: equation w. radicals and exponenets

    thanks so much prove it, but can you explain when you get the (x-7) squared, how do you elimate the second (x-7) in the last step? does it cancel itself out? dont understand that step, thanks again
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    Re: equation w. radicals and exponenets

    Quote Originally Posted by mathmathmathmathmathmathm View Post
    thanks so much prove it, but can you explain when you get the (x-7) squared, how do you elimate the second (x-7) in the last step? does it cancel itself out? dont understand that step, thanks again
    To undo a square, you need to square root both sides. The square root of 0 is 0.
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