
Compute the values
for 3X4Y
knowing that X=2 AND Y=1 and X.Y=1
and then compute for
(3X+4Y).(X5Y)
apparently the answer for the first is square root of 76
and the second is 1
I tried doing this the way the book tells me but I don't get the right answer at ALL!

$\displaystyle \begin{array}{rcl}
\left\ {3X  4Y} \right\^2 & = & \left( {3X  4Y} \right) \cdot \left( {3X  4Y} \right) \\
& = & 9\left\ X \right\^2 + 16\left\ Y \right\^2  24\left( {X \cdot Y} \right) \\ \end{array}$
Just make the substitutions and the squareroot of both sides.