If $m, n$ are the roots of the equation $K (x^{2} - x) + x + 5 = 0.$ If $K_{1} and K_{2}$ are the two values of $K$ for which the roots $m, n$ are connected by the relation $\frac{m}{n} + \frac{n}{m} = \frac{4}{5} .$ Then the value of $\frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}}$ is
254

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Solution

The correct option is A 254
Given: $K (x^{2} - x) + x + 5 = 0,$ roots are $m, n$ and $\frac{m}{n} + \frac{n}{m} = \frac{4}{5} .$

We have, $K (x^{2} - x) + x + 5 = 0$
$\Rightarrow K x^{2} + (1 - K) x + 5 = 0$

$\Rightarrow$ Sum of roots $= m + n = \frac{K - 1}{K} \dots (i)$

$\Rightarrow$ Product of roots $= m . n = \frac{5}{K} \dots (i i)$

Also we have, $\frac{m}{n} + \frac{n}{m} = \frac{4}{5}$
$\Rightarrow \frac{m^{2} + n^{2}}{m n} = \frac{4}{5}$

$\Rightarrow \frac{(m + n)^{2} - 2 m n}{m n} = \frac{4}{5}$

$\Rightarrow \frac{(\frac{K - 1}{K})^{2} - 2 \frac{5}{K}}{\frac{5}{K}} = \frac{4}{5} [From (i) and (i i)]$

$\Rightarrow \frac{K^{2} - 12 K + 1}{5 K} = \frac{4}{5}$

$\Rightarrow K^{2} - 16 K + 1 = 0$

$\Rightarrow$ Sum of roots $= K_{1} + K_{2} = 16 \dots (i i i)$

$\Rightarrow$ Product of roots $= K_{1} . K_{2} = 1 \dots (i v)$

We have, $\frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}}$

$\frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}} = \frac{K_{1}^{2} + K_{2}^{2}}{K_{1} . K_{2}}$

$\Rightarrow \frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}} = \frac{(K_{1} + K_{2})^{2} - 2 K_{1} . K_{2}}{K_{1} . K_{2}}$

$\Rightarrow \frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}} = \frac{(16)^{2} - 2.1}{1} [From (i i i) and (i v)]$

$\Rightarrow \frac{K_{1}}{K_{2}} + \frac{K_{2}}{K_{1}} = 254$

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