Simplifying Error

The quantity given is negative while all of the numerical answer choices are positive.

Challenge master's bonus question solution:

Let's assume that $r$ is the root of such polynomial,hence:

$r=\sqrt3 - \sqrt5 \rightarrow r^2 = 8 -2\sqrt15 \rightarrow r^2-8=-2\sqrt15$

$r^4+64-16r^2=30 \rightarrow r^4-16r^2+34=0$

Hence the equation $x^4-16x^2+34=0$ has at least one root which is $\sqrt3 - \sqrt5$

And of course it's degree is 4.

$\quad$

As for our dear challenge master's second bonus question:

By false assumption , suppose that $\sqrt{3} - \sqrt{5}$ is rational.Hence it can be written the form of $\frac{m}{n}$ where $m$ and $n$ are coprime integers.Now we know that if $gcd(m,n)=1 \rightarrow gcd(m^2,n^2)=1$ hence , ,let $\frac{m^2}{n^2} = \frac{p}{q}$ where $gcd (p,q)=1$ thus, we have:

$\quad$

$\frac{p^2}{q^2}-16 \frac{p}{q}=-34 \rightarrow \frac{p^2q-16pq^2}{q^2}=-34$

$\rightarrow \frac{(pq)(p-16q)}{q^2} = -34 \rightarrow \frac{(p)(p-16q)} {q} = -34$

$\rightarrow q|p \lor q|p-16q$

$\quad$

Case 1:if $q|p$

Directly contradicts our assumption of $gcd(p,q)=1$

Case 2: if $q|p-16q$

$q|p-16q$

$q|16q$

$\Rightarrow q|p-16q+16q \rightarrow q|p$

Which is again also a contradiction.

$\quad$

Hence $\sqrt{3} - \sqrt{5}$ is indeed irrational.

What I did was square the entire term. That's it.

Right answer: square root of 3 - square root of 5