A store manager is very impatient at the start of the day. Let $X$ be the time (in minutes) until his first customer arrives. The workers at the store decided to find a way to measure his impatience level as a function of $X$ where $X$ belongs to an exponential distribution with
$g(X) = aX^2 + bX + c$
a. Calculate the manager’s expected impatience level. $E(g(X)) = aE(X^2) + bE(X) + c = \cfrac{2a}{\lambda^2} + \cfrac{b}{\lambda} + c$
b. Find the variance of the manager’s impatience level.
$Var(g(X)) =a^2Var(X^2) + b^2Var(X) = \cfrac{a^220}{\lambda^4} + \cfrac{b^2}{\lambda^2}$
But the answer for b) is $Var(g(X)) = \cfrac{b^2}{\lambda^2} + \cfrac{8ab}{\lambda^3} + \cfrac{20a^2}{\lambda^4}$
What is my error in part b)?
You forgot the covariance, you can't omit it because the random variables aren't independent, then
$$Var(g(x)) = a^2Var(X^2)+2ab \ \text{Cov}(X^2,X)+b^2Var(X)$$
where $$\text{Cov}(X^2,X)=E(X^3)-E(X)E(X^2) = \frac{3!}{\lambda^3}-\frac{2}{\lambda^3} = \frac{4}{\lambda^3}$$
So you get
$$Var(g(x)) = \cfrac{b^2}{\lambda^2} + \cfrac{8ab}{\lambda^3} + \cfrac{20a^2}{\lambda^4}$$