hi, i can't sort out what to do with this one:

$\displaystyle

\sqrt{2m-3}=\sqrt{m+7} - 2

$

thanks

jacs

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- Apr 9th 2006, 05:05 AMjacsEquation
hi, i can't sort out what to do with this one:

$\displaystyle

\sqrt{2m-3}=\sqrt{m+7} - 2

$

thanks

jacs - Apr 9th 2006, 05:38 AMOReilly
$\displaystyle \begin{array}{l}

\sqrt {2m - 3} = \sqrt {m + 7} - 2 \\

\left( {\sqrt {2m - 3} } \right)^2 = \left( {\sqrt {m + 7} - 2} \right)^2 \\

2m - 3 = m + 7 - 4\sqrt {m + 7} + 4 \\

2m - 3 - m - 7 - 4 = - 4\sqrt {m + 7} \\

\end{array}

$

$\displaystyle \begin{array}{l}

m - 14 = - 4\sqrt {m + 7} \\

\left( {m - 14} \right)^2 = \left( {4\sqrt {m + 7} } \right)^2 \\

m^2 - 28m + 196 = 16(m + 7) \\

m^2 - 28m + 196 - 16m - 112 = 0 \\

m^2 - 44m + 84 = 0 \\

\end{array}

$

$\displaystyle m_{1/2} = \frac{{44 \pm \sqrt {1936 - 336} }}{2} = \frac{{44 \pm \sqrt {1600} }}{2} = \frac{{44 \pm 40}}{2}$

$\displaystyle m_1 = 42 \wedge m_2 = 2$

Solution is $\displaystyle m=2$. - Apr 9th 2006, 06:14 AMjacs
thanks OReilly, i see it now. I got half way and then didnt know how to proceed, but it is just manipulating them cleverly.

thanks,for that

i appreciate it

jacs