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

• A. 7
• B. 8
• C. 9
• D. None of Above

Answer 1

The Correct Option is C

let \(\alpha \& \beta\) be the roots of \(x^2+bx+c=0\)
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5}{9}\)
\(\left( \frac{1}{\alpha} - \frac{1}{\beta} \right)^2 = \frac{1}{9}\)

\(\frac{4}{9} = \frac{2}{\alpha \beta}\)

\(\alpha \beta = \frac{9}{2} \quad \alpha^2 + \beta^2 = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = \frac{5}{9}\)

\((\alpha + \beta)^2 = \frac{5}{9} \left( \frac{9}{2} \right) \left( \frac{9}{2} \right) + 2 \left( \frac{9}{2} \right)\)

\((\alpha + \beta)^2 = \frac{45}{4} + 9 = 20 + 14 = 20.25\)

\(\alpha + \beta = \pm 4.5\)

\(\alpha + \beta = -\frac{b}{1}\)

\(b=−(α+β)\)

\(\alpha \beta = \frac{c}{1}\)

c=αβ=4.5

\(b + c = -(\alpha + \beta) + \alpha\)
to maximize\((b + c), \text{we pick } (\alpha + \beta) = -4.5\)
\(b+c=−(−4.5)+4.5=9\)
The correct option is (C): 9.