For some real number $r$, the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$. Which of the following numbers is closest to $r$? 
$\text{(A)} \ 1.22 \qquad  \text{(B)} \ 1.32 \qquad  \text{(C)} \ 1.42 \qquad  \text{(D)} \ 1.52 \qquad  \text{(E)} \ 1.62$
Solution by e_power_pi_times_i
Denote $s$ as the third solution. Then, by Vieta's, $2r+s = \dfrac{1}{2}$, $r^2+2rs = -\dfrac{21}{4}$, and $r^2s = -\dfrac{45}{8}$. Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\dfrac{21}{4}$. Combined with the fact that $s = \dfrac{1}{2}-2r$, the third equation can be written as $(\dfrac{r+\dfrac{21}{4}}{3})(\dfrac{1}{2}-2r) = -\dfrac{45}{8}$, or $(4r+21)(4r-1) = 135$. Solving, we get $r = \dfrac{3}{2}, -\dfrac{13}{2}$. Plugging the solutions back in, we see that $-\dfrac{13}{2}$ is an extraneous solution, and thus the answer is $\boxed{\text{(D)} \ 1.52}$