Question

If k is ratio of zeroes of the polynomial $4x^2-5x-3,$ then the value of ${(k+\frac{1}{k})}^{-2}$ is

• A. $\frac{49}{12}$
• B. $\frac{12}{49}$
• C. $\frac{144}{2401}$
• D. $\frac{2401}{144}$

Answer 1

The correct option is C $\frac{144}{2401}$

Given polynomial is $4x^2-5x-3,$ and k is the ratio of its zeroes.

Let $\alpha and\beta$ be the zeroes of the polynomial, then $k=\frac{\alpha }{\beta }.$

Also, $\alpha +\beta =\frac{-(-5)}{4}=\frac{5}{4}$

and, $\alpha \beta =\frac{-3}{4}$

Now, $k+\frac{1}{k}=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$ $(\because k=\frac{\alpha }{\beta })$

$=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{{\left(\frac{5}{4}\right)}^2-2\times \left(\frac{-3}{4}\right)}{\left(\frac{-3}{4}\right)}(\because \alpha +\beta =\frac{5}{4};\alpha \beta =\frac{-3}{4})$

$=\frac{\frac{25}{16}+\frac{3}{2}}{\left(\frac{-3}{4}\right)}$

$=\frac{-49}{12}$

$\therefore {(k+\frac{1}{k})}^{-2}={\left(\frac{-49}{12}\right)}^{-2}$

$={\left(\frac{-12}{49}\right)}^2$

$=\frac{144}{2401}$

Thus, the value of ${(k+\frac{1}{k})}^{-2}is\frac{144}{2401}.$

Hence, the correct answer is option (c).