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CAT 2023 Slot 3 — QA Question 6

Forming a Quadratic Equation and Relation between roots and coefficientsEasy

A quadratic equation x2 + bx + c = 0 has two real roots. If the difference between the reciprocals of the roots is 1/3, and the sum of the reciprocals of the squares of the roots is 5/9, then the largest possible value of (b + c) is

Answer & solution

Answer: 9

Solution

Easy

Let the roots be p,qp,q. By Vieta, p+q=−bp+q=-b and pq=cpq=c. Convert the two given conditions (difference of reciprocals =13=\tfrac13, sum of reciprocal-squares =59=\tfrac59) into equations in p+qp+q and pqpq, solve for both, and maximise b+cb+c.

1

Write the two conditions:

1p−1q=13(1),1p2+1q2=59(2)\frac1p-\frac1q=\frac13\quad(1),\qquad \frac1{p^2}+\frac1{q^2}=\frac59\quad(2)
2

Square (1) and substitute (2) to find pqpq:

1p2+1q2−2pq=19⇒ 59−2pq=19⇒ 2pq=49 ⇒ pq=92=c\begin{aligned} &\frac1{p^2}+\frac1{q^2}-\frac{2}{pq}=\frac19\\ &\Rightarrow\ \frac59-\frac{2}{pq}=\frac19\\ &\Rightarrow\ \frac{2}{pq}=\frac49\ \Rightarrow\ pq=\frac92=c \end{aligned}
3

Use (2) to find p+qp+q. Multiply through by p2q2p^2q^2:

9(p2+q2)=5(pq)2⇒ 9((p+q)2−2pq)=5(92)2⇒ (p+q)2−9=454⇒ (p+q)2=814 ⇒ p+q=±92\begin{aligned} &9(p^2+q^2)=5(pq)^2\\ &\Rightarrow\ 9\big((p+q)^2-2pq\big)=5\left(\tfrac92\right)^2\\ &\Rightarrow\ (p+q)^2-9=\frac{45}{4}\\ &\Rightarrow\ (p+q)^2=\frac{81}{4}\ \Rightarrow\ p+q=\pm\frac92 \end{aligned}
4

Maximise b+cb+c. Since b=−(p+q)b=-(p+q), the largest bb comes from p+q=−92p+q=-\tfrac92, giving b=92b=\tfrac92. With c=92c=\tfrac92:

b+c=92+92=9b+c=\frac92+\frac92=9

Largest possible value of b+c=9b+c=\mathbf{9}.

Standard symmetric-function moves: 1p2+1q2=p2+q2(pq)2\dfrac1{p^2}+\dfrac1{q^2}=\dfrac{p^2+q^2}{(pq)^2} and p2+q2=(p+q)2−2pqp^2+q^2=(p+q)^2-2pq turn both conditions into equations in just p+qp+q and pqpq.

CAT 2023 Slot 3 QA Q6: A quadratic equation x 2 + bx + c = 0 has two real roots. If the difference between the reciprocals of the roo — Solution | TheCATExam