CAT 2023 Slot 3 QA Question & Solution
Question
A quadratic equation $x^2 + bx + c = 0$ has two real roots. If the difference between the reciprocals of the roots is $\frac{1}{3}$, and the sum of the reciprocals of the squares of the roots is $\frac{5}{9}$, then the largest possible value of $(b + c)$ is
Solution
It is given that $x^2 + bx + c = 0$ has two real roots. Let the roots of the equation be $\alpha,\ \beta$ ($\alpha > \beta$)
Then, we can say that $\frac{1}{\alpha\ }-\frac{1}{\beta\ }=\frac{1}{3}$ .... Eq(1)
Similarly, $\frac{1}{\alpha\ ^2}+\frac{1}{\beta^2\ }=\frac{5}{9}$ .... Eq (2)
Eq(2) can be written as $\left(\frac{1}{\alpha\ }-\frac{1}{\beta\ }\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$
=> $\left(\frac{1}{3}\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$
=> $\frac{2}{\alpha\ \cdot\beta\ }=\frac{4}{9}=>\ \frac{1}{\alpha\ \cdot\beta\ }=\frac{2}{9}$
=> $\alpha\ \cdot\beta\ =\frac{9}{2}$
We know that the product of the roots is equal to c, which implies$c=\frac{9}{2}$
We also know that the sum of the roots is equal to -b.
=> $\frac{1}{\alpha\ ^2}+\frac{1}{\beta\ ^2}=\left(\frac{1}{\alpha\ }+\frac{1}{\beta\ }\right)^{^2}-\frac{2}{\alpha\ \beta\ }=\frac{5}{9}$
=> $\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}-\frac{4}{9}=\frac{5}{9}$
=> $\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}=\left(1\right)^2$
\(\alpha + \beta = \pm \alpha\beta\)
Hence, the maximum value of b is $\frac{9}{2}$.
Hence, the maximum value of (b+c) is 9