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Math Help - Another radical equation question

  1. #1
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    Another radical equation question

    Solve the equation. Check for extraneous solutions.

    <br />
\sqrt{2x + 3} + 2 = \sqrt{6x + 7}<br />

    I get to

    <br />
0 = 4x<br />

    Does that mean no solution?
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  2. #2
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    Quote Originally Posted by happydino1 View Post
    Solve the equation. Check for extraneous solutions.

    <br />
\sqrt{2x + 3} + 2 = \sqrt{6x + 7}<br />

    I get to

    <br />
0 = 4x<br />

    Does that mean no solution?
    \left( \sqrt{2x+3}+2 \right)^2 = 6x+7
    2x+3 + 4 + 4\sqrt{2x+3} = 6x+7
    4\sqrt{2x+3} = 4x
    \sqrt{2x+3}=x
    2x+3 = x^2
    Can you solve that?
    Last edited by ThePerfectHacker; December 31st 2007 at 04:00 PM.
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