A &#x2282;<!-- ⊂ --> B is a ring extension. Let y , z &#x2208;<!-- ∈ -->

Follow the idea of the next exercise or you may want to start with writing
$y^2+ay+b=0⟹y=\frac{-a-\sqrt{a^2-4b}}{2}\text{ or }y=\frac{-a+\sqrt{a^2-4b}}{2}.$
For a while choose $y=\frac{-a-\sqrt{a^2-4b}}{2}$. Same for $z^2+cz+d=0$.
Then $y+z=\frac{-(a+c)-(\sqrt{a^2-4b}+\sqrt{c^2-4d})}{2}$
$⟹(y+z)+\frac{a+c}{2}=-\frac{\sqrt{a^2-4b}+\sqrt{c^2-4d}}{2}$
By first, squaring both sides you will earn a monic relation for $y+z$ but it has (only) one sqrt so after squaring it for second time you will have a monic relation with no sqrt at $(A[\frac{1}{2}])[T]$. If $\frac{1}{2}\in A$ then everything is finished without needing computing the exact relations but if you simplify it, no coefficient will have $\frac{1}{2}$.
The same method will work for $xy$ but I have doubt if the second method give us a better monic relation than a relation at $(A[\frac{1}{2}])[T]$, you may want to compute it to see.
After making first square you will have;
$(y+z{)}^2+(a+c)(y+z)+\frac{(a+c{)}^2}{4}=\frac{a^2-4b+c^2+2\sqrt{a^2-4b}\sqrt{c^2-4d}}{4}$
$(y+z{)}^2+(a+c)(y+z)+\frac{2ac}{4}=-b-d+\frac{\sqrt{a^2-4b}\sqrt{c^2-4d}}{2}$
Now we go for the second squaring, but as we want to get rid of radicals we take $-b-d$ to the left hand side and then we square sides.
$(y+z{)}^4+2(a+c)(y+z{)}^3+2(\frac{ac}{2}+b+d)(y+z{)}^2+2(a+c)(\frac{ac}{2}+b+d)(y+z)+(\frac{ac}{2}+b+d{)}^2=\frac{(a^2-4b)(c^2-4d)}{4}$
It's obvious that all coefficients will be in $A$ and won't have $\frac{1}{2}$ and we only need to pay attention to the constant coefficient.
$(\frac{a^2{c}^2}{4}+b^2+d^2+acb+acd+2bd)-(\frac{a^2{c}^2}{4}-a^{2}d-b{c}^2+4bd)$
Now one can see the simplified form and for sure there is no fractions like $\frac{1}{2}$ or $\frac{1}{4}$.
"For $xy$ we will encounter 3 radicals! What can we do?"
Don't be afraid! 3 radicals is not a very scay case yet. For getting rid of 3 radicals in an equation like $a+\sqrt{b}+\sqrt{c}+\sqrt{d}=0$ do as following;
$a+\sqrt{b}+\sqrt{c}+\sqrt{d}=0⟹a+\sqrt{b}=-(\sqrt{c}+\sqrt{d})$
Take a square
$a^2+2a\sqrt{b}+b=c+d+2\sqrt{cd}$
Then we are in case with two radicals;
$a^2+b-c-d=2\sqrt{b}+2\sqrt{cd}$
Now let's be sure that we won't have coefficients with $\frac{1}{2}$ and so on.
$xy=\frac{a+\sqrt{a^2-4b}}{2}\frac{c^2+\sqrt{c^2-4d}}{2}=\frac{ac+a\sqrt{c^2-4d}+c\sqrt{a^2-4b}+\sqrt{a^2-4b}\sqrt{c^2-4d}}{4}$
$xy-\frac{ac}{4}-\frac{\sqrt{(a^2-4b)(c^2-4d)}}{4}=\frac{a}{4}\sqrt{c^2-4d}+\frac{c}{4}\sqrt{a^2-4b}$
$(xy{)}^2+\frac{a^2{c}^2}{16}+\frac{(a^2-4b)(c^2-4d)}{16}-\frac{ac}{2}(xy)-\frac{(a^2-4b)(c^2-4d)}{2}((xy)-\frac{ac}{4})=\frac{a^2}{16}(c^2-4d)+\frac{c^2}{16}(a^2-4b)+\frac{ac}{8}\sqrt{(c^2-4d)(a^2-4b)}$
$(xy{)}^2+\frac{a^2{c}^2}{16}+\frac{a^2{c}^2}{16}-\frac{a^{2}d}{4}-\frac{b{c}^2}{4}+bd-\frac{ac}{2}(xy)-\frac{a^2{c}^2}{16}+\frac{a^{2}d}{4}-\frac{c^2{a}^2}{16}+\frac{b{x}^2}{4}=(\frac{ac}{8}+\frac{1}{2}(xy)-\frac{ac}{8})\sqrt{(c^2-4d)(a^2-4b)}$
$(xy{)}^2-\frac{ac}{2}(xy)+bd=\frac{1}{2}(xy)\sqrt{(c^2-4d)(a^2-4b)}$
$(xy{)}^4-ac(xy{)}^3-\frac{a^2{c}^2}{4}(xy{)}^2+2bd(xy{)}^2-acbd(xy)+b^2{d}^2=\frac{1}{4}(xy{)}^2(c^2-4d)(a^2-4b)$
And in the end
$(xy{)}^4-ac(xy{)}^3+(c^{2}b+d{a}^2-2bd)(xy{)}^2-acbd(xy)+b^2{d}^2=0$