Question 51

# 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

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