Answer
{$\frac{3-\sqrt {137}}{8},\frac{3+\sqrt {137}}{8}$}
Work Step by Step
First, we subtract the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$:
$n-\frac{2}{n}=\frac{3}{4}$
$\frac{n(n)-2(1)}{n}=\frac{3}{4}$
$\frac{n^{2}-2}{n}=\frac{3}{4}$
Now, we cross multiply the two fractions in order to create a linear equation:
$\frac{n^{2}-2}{n}=\frac{3}{4}$
$4(n^{2}-2)=3n$
$4n^{2}-8=3n$
$4n^{2}-3n-8=0$
Now, we will use the quadratic formula to solve the equation:
Step 1: Comparing $4n^{2}-3n-8=0$ to the standard form of a quadratic equation, $an^{2}+bn+c=0$, we find:
$a=4$, $b=-3$ and $c=-8$
Step 2: The quadratic formula is:
$n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$n=\frac{-(-3) \pm \sqrt {(-3)^{2}-4(4)(-8)}}{2(4)}$
Step 4: $n=\frac{3 \pm \sqrt {9+128}}{8}$
Step 5: $n=\frac{3 \pm \sqrt {137}}{8}$
Step 6: $n=\frac{3-\sqrt {137}}{8}$ or $n=\frac{3+\sqrt {137}}{8}$
Step 7: Therefore, the solution set is {$\frac{3-\sqrt {137}}{8},\frac{3+\sqrt {137}}{8}$}.
