Find all values of $a$ that satisfy the equation \[\frac{a}{3} + 1 = \frac{a + 3}{a} - \frac{a^2 + 2}{a}.\]
Let's simplify:
\(\frac{a}{3}+1=1+\frac{3}{a}-a+\frac{2}{a}\)
\(a-\frac{5}{a}+\frac{a}{3}=0\)
I just put this on a calculator and got a = \(\frac{\sqrt 15}{2}\), \(\frac{-\sqrt 15}{2}\)
I didn't know what to do.
Answer: (sqrt 15)/2, -(sqrt 15)/2
Simplify as
(a + 3) / 3 = [ -a^2 + a + 1 ] / a cross-multiply
a ( a + 3) = 3 (-a^2 + a + 1)
a^2 + 3a = -3a^2 + 3a + 3
4a^2 = 3
a^2 = 3/4 take both roots
a = sqrt (3) / 2 or a = -sqrt (3) / 2
+1364 
