# Equation

• Apr 9th 2006, 05:05 AM
jacs
Equation
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 AM
OReilly
$\displaystyle \begin{array}{l} \sqrt {2m - 3} = \sqrt {m + 7} - 2 \\ \left( {\sqrt {2m - 3} } \right)^2 = \left( {\sqrt {m + 7} - 2} \right)^2 \\ 2m - 3 = m + 7 - 4\sqrt {m + 7} + 4 \\ 2m - 3 - m - 7 - 4 = - 4\sqrt {m + 7} \\ \end{array}$
$\displaystyle \begin{array}{l} m - 14 = - 4\sqrt {m + 7} \\ \left( {m - 14} \right)^2 = \left( {4\sqrt {m + 7} } \right)^2 \\ m^2 - 28m + 196 = 16(m + 7) \\ m^2 - 28m + 196 - 16m - 112 = 0 \\ m^2 - 44m + 84 = 0 \\ \end{array}$
$\displaystyle m_{1/2} = \frac{{44 \pm \sqrt {1936 - 336} }}{2} = \frac{{44 \pm \sqrt {1600} }}{2} = \frac{{44 \pm 40}}{2}$
$\displaystyle m_1 = 42 \wedge m_2 = 2$

Solution is $\displaystyle m=2$.
• Apr 9th 2006, 06:14 AM
jacs
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