CAT 2022 Slot 3 · Quant · Question 14

When a quadratic has integer coefficients and an irrational root of the form p+q\sqrt{d}, its conjugate p-q\sqrt{d} is the other root.

So the first equation has roots 3+2\sqrt{2} and 3-2\sqrt{2}.

Their sum is 6, and Vieta gives -\frac{b}{a}=6, so b=-6a.

Their product is 9-8=1, and Vieta gives \frac{c}{a}=1, so c=a.

The second equation has roots 4+2\sqrt{3} and 4-2\sqrt{3}, and leading coefficient a as well.

Their sum is 8, so -\frac{m}{a}=8 and m=-8a.

Their product is 16-12=4, so \frac{n}{a}=4 and n=4a.

Now evaluate the first fraction: \frac{b}{m}=\frac{-6a}{-8a}=\frac{3}{4}.

For the second, c-2b=a-2\left( -6a \right)=a+12a=13a.

So \frac{c-2b}{n}=\frac{13a}{4a}=\frac{13}{4}.

Adding, \frac{3}{4}+\frac{13}{4}=\frac{16}{4}=4 — the third option.

Shortcut. Every coefficient is a multiple of a, so a cancels from both fractions and the answer is a pure number. Setting a=1 from the start gives b=-6, c=1, m=-8 and n=4 with no symbols to carry.