How do I get to this answer without have to solve for a 4th degree polynomial?

Question

This is question regarding solving for a percentile of the cumulative distribution function of a certain pdf. What I am wondering is how I am meant to get to this answer without having to solve a complex polynomial. Any help is much appreciated!!

Answer 1

Hint

Let $(\pi_{0.75}+6)^2=x$ and solve for $x$ $$\frac 3 4=1-\left(1-\frac x {60} +\frac 4 {60}\right)^2$$ that is to say $$\frac 1 4=\left(\frac{16}{15}-\frac{x}{60}\right)^2$$ which gives as solutions $x=34$ and $x=94$.

This is not too complex.

I am sure that you can take it from here.

Answer 2

\begin{align*} \dfrac{3}{4} &= 1 - \left[1-\dfrac{1}{60}(\pi_{0.75}+6)^2 +\frac{4}{60} \right]^2 \\ \dfrac{-1}{4} &= - \left[1-\dfrac{1}{60}(\pi_{0.75}+6)^2 +\frac{4}{60} \right]^2 \\ \dfrac{1}{4} &= \left[1-\dfrac{1}{60}(\pi_{0.75}+6)^2 +\frac{4}{60} \right]^2 \\ \pm \dfrac{1}{2} &= 1-\dfrac{1}{60}(\pi_{0.75}+6)^2 +\frac{4}{60} \\ \pm \dfrac{30}{60} - \dfrac{60}{60} - \frac{4}{60} &= -\dfrac{1}{60}(\pi_{0.75}+6)^2 \\ \dfrac{\{-34,-94\}}{60} &= -\dfrac{1}{60}(\pi_{0.75}+6)^2 \\ \{34,94\} &= (\pi_{0.75}+6)^2 \\ \{\pm\sqrt{34},\pm\sqrt{94}\} &= \pi_{0.75}+6 \\ \{\pm\sqrt{34} - 6,\pm\sqrt{94} - 6\} &= \pi_{0.75} \text{,} \end{align*} where we have used the somewhat nonstandard notation "$\text{set} = \text{expression}$" ro represent that the expression can take any value in the set.

This gives all four roots of your polynomial.