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Quantitative Aptitude Question on Square and Square Roots

A quadratic equation x2+bx+c=0x^2+bx+c=0 has two real roots. If the difference between the reciprocals of the roots is 13\frac{1}{3} , and the sum of the reciprocals of the squares of the roots is 59\frac{5}{9} , then the largest possible value of (b+c)(b+c) is

A

7

B

8

C

9

D

None of Above

Answer

9

Explanation

Solution

let α&β\alpha \& \beta be the roots of x2+bx+c=0x^2+bx+c=0
1α2+1β2=59\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5}{9}
(1α−1β)2=19\left( \frac{1}{\alpha} - \frac{1}{\beta} \right)^2 = \frac{1}{9}

49=2αβ\frac{4}{9} = \frac{2}{\alpha \beta}

αβ=92α2+β2=α2+β2α2β2=(α+β)2−2αβ(αβ)2=59\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}

(α+β)2=59(92)(92)+2(92)(\alpha + \beta)^2 = \frac{5}{9} \left( \frac{9}{2} \right) \left( \frac{9}{2} \right) + 2 \left( \frac{9}{2} \right)

(α+β)2=454+9=20+14=20.25(\alpha + \beta)^2 = \frac{45}{4} + 9 = 20 + 14 = 20.25

α+β=±4.5\alpha + \beta = \pm 4.5

α+β=−b1\alpha + \beta = -\frac{b}{1}

b=−(α+β)b=−(α+β)

αβ=c1\alpha \beta = \frac{c}{1}

c=αβ=4.5

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