CAT 2023 Slot 3 · Quant · Question 5

Let the roots be p and q, so p+q=-b and pq=c.

The two conditions involve reciprocals, so build the identity that links them.

Squaring the difference, {\left( \frac{1}{p}-\frac{1}{q}\right)}^{2}=\frac{1}{{p}^{2}}+\frac{1}{{q}^{2}}-\frac{2}{pq}.

Substituting the given values, \frac{1}{9}=\frac{5}{9}-\frac{2}{pq}.

So \frac{2}{pq}=\frac{4}{9}, giving pq=\frac{9}{2}, that is c=\frac{9}{2}.

Now the matching sum: {\left( \frac{1}{p}+\frac{1}{q}\right)}^{2}=\frac{1}{{p}^{2}}+\frac{1}{{q}^{2}}+\frac{2}{pq}=\frac{5}{9}+\frac{4}{9}=1.

So \frac{1}{p}+\frac{1}{q}=\pm 1.

But \frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}, so p+q=\pm \frac{9}{2} and therefore b=\mp \frac{9}{2}.

To make b+c largest, take b=\frac{9}{2}, which pairs with p+q=-\frac{9}{2}.

Then b+c=\frac{9}{2}+\frac{9}{2}=9.

Checking the roots are real: the discriminant is {b}^{2}-4c=\frac{81}{4}-18=\frac{9}{4}>0, so both roots are real as required.

Shortcut. Adding and subtracting the squares of the reciprocal sum and difference is the standard move, and it delivers pq and p+q in two lines. Since only \left| p+q \right| is pinned down, the sign is free — pick the one that makes b positive.