Solve the equation. Check for extraneous solutions. $\displaystyle \sqrt{2x + 3} + 2 = \sqrt{6x + 7} $ I get to $\displaystyle 0 = 4x $ Does that mean no solution?
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Originally Posted by happydino1 Solve the equation. Check for extraneous solutions. $\displaystyle \sqrt{2x + 3} + 2 = \sqrt{6x + 7} $ I get to $\displaystyle 0 = 4x $ Does that mean no solution? $\displaystyle \left( \sqrt{2x+3}+2 \right)^2 = 6x+7$ $\displaystyle 2x+3 + 4 + 4\sqrt{2x+3} = 6x+7$ $\displaystyle 4\sqrt{2x+3} = 4x $ $\displaystyle \sqrt{2x+3}=x$ $\displaystyle 2x+3 = x^2$ Can you solve that?
Last edited by ThePerfectHacker; Dec 31st 2007 at 04:00 PM.
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