CAT 2023 Slot 1 · Quant · Question 6

From 2{x}^{2}-6x+k=0, the sum of the roots is \alpha +\beta =\frac{6}{2}=3 and the product is \alpha \beta =\frac{k}{2}.

These two numbers, 3 and \frac{k}{2}, are stated to be the roots of {x}^{2}+px+p=0.

For that second equation, the sum of its roots is -p and the product is p.

Matching sums, 3+\frac{k}{2}=-p.

Matching products, 3\times \frac{k}{2}=p, so p=\frac{3k}{2}.

Substituting into the sum equation, 3+\frac{k}{2}=-\frac{3k}{2}.

Collecting the k terms, 3=-\frac{3k}{2}-\frac{k}{2}=-2k.

So k=-\frac{3}{2}, and then p=\frac{3}{2}\times \left( -\frac{3}{2}\right)=-\frac{9}{4}.

Now k-p=-\frac{3}{2}+\frac{9}{4}=\frac{-6+9}{4}=\frac{3}{4}.

Therefore 8\left( k-p \right)=8\times \frac{3}{4}=6.

Shortcut. The factor 8 in the question is the clue that k and p will come out in quarters and halves. Since the sum of the roots of the first equation is a plain 3, the only unknown carried forward is k, and the two Vieta relations of the second equation pin it down in two lines.