if 1/a + 1/b = 3/8 and a*b = 3,
then find a+b and a^2 + b^2
HELP??
First:
$\displaystyle
\frac{1}{a}+\frac{1}{b}=\frac{b+a}{ab}=\frac{a+b}{ 3}=\frac{3}{8}
$
Hence:
$\displaystyle a+b=\frac{9}{8}$
Second:
$\displaystyle (a+b)^2=a^2+b^2+2ab=(a^2+b^2)+6$
rearranging:
$\displaystyle
a^2+b^2=6-(a+b)^2=6-\left[\frac{9}{8}\right]^2=\frac{303}{64}
$
RonL