Question

If $\alpha ,\beta$ are the roots of the equation $k(x^2-x)+x+5=0$. If $k_1\&k_2$ are the two values of k for which the roots $\alpha ,\beta$ are connected by the relation $\left(\alpha /\beta \right)+\left(\beta /\alpha \right)=4/5$. Find the value of $\left(k_1/k_2\right)+\left(k_2/k_1\right)$.

Answer 1

Given equation is $k(x^2-x)+x+5=0$

$⟹k{x}^2+(1-k)x+5=0$

Given that $\alpha ,\beta$ are the roots of the above equation.

Therefore, sum of the roots is $\alpha +\beta =\frac{-(1-k)}{k}$ .......(1)

and the product of the roots is $\alpha \beta =\frac{5}{k}$ ......(2)

Given that $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{5}$

$⟹\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{4}{5}$

$⟹{\alpha }^2+{\beta }^2=\frac{4\alpha \beta }{5}$

$⟹(\alpha +\beta {)}^2-2\alpha \beta =\frac{4\alpha \beta }{5}$ .....(3)

substituting (1) and (2) in (3) we get

$(\frac{k-1}{k}{)}^2-2(\frac{5}{k})=\frac{4\left(\frac{5}{k}\right)}{5}$

$⟹\frac{(k-1{)}^2}{k^2}-\frac{10k}{k^2}=\frac{4k}{k^2}$

$⟹(k-1{)}^2-10k=4k$

$⟹k^2+1-2k-14k=$

$⟹k^2-16k+1=$

$⟹k=\frac{16\pm \sqrt{{16}^2-4}}{2}$

$⟹k=\frac{16\pm \sqrt{252}}{2}$

$⟹k=8\pm \sqrt{63}$

Given that $k_1,k_2$ are the two values of $k$

Therefore, $k_1=8+\sqrt{63},k_2=8-\sqrt{63}$

Hence, $\frac{k_1}{k_2}+\frac{k_2}{k_1}=\frac{8+\sqrt{63}}{8-\sqrt{63}}+\frac{8-\sqrt{63}}{8+\sqrt{63}}$

$=\frac{(8+\sqrt{63}{)}^2+(8-\sqrt{63}{)}^2}{(8+\sqrt{63})(8-\sqrt{63})}$

$=\frac{64+63+16\sqrt{63}+64+63-16\sqrt{63}}{64-63}$

$=\frac{254}{1}=254$