Question

Given $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+k=0(k\ne 0)$. If $\alpha \beta ,\alpha {\beta }^2+{\alpha }^2\beta ,{\alpha }^3+{\beta }^3$ are in geometric progression then the value of ${}^{\prime }{k}^{\prime }$ equals

• A. $4$
• B. $\frac{16}{7}$
• C. $\frac{3}{7}$
• D. $12$

Answer 1

The correct option is B $\frac{16}{7}$

If $a{x}^2+bx+c=0$ is a quadratic equation, then the relation between it's roots is given as :

Sum of the roots : $\alpha +\beta =\frac{-b}{a}$

Product of the roots: $\alpha \beta =\frac{c}{a}$

$\begin{array}{c}{x}^2-4x+k=0\\ \alpha +\beta =4\\ \alpha \beta =k\\ \alpha \beta ,\alpha {\beta }^2+{\alpha }^2\beta ,{\alpha }^3+{\beta }^3,are,in,GP\\ \Rightarrow {\left(\alpha {\beta }^2+{\alpha }^2\beta \right)}^2=\alpha \beta \left({\alpha }^3+{\beta }^3\right)\\ \Rightarrow \left[\alpha \beta \right(\alpha +\beta ){]}^2=\alpha \beta (\alpha +\beta \left)\right({\alpha }^2-\beta \alpha +{\beta }^2)\\ \Rightarrow k^2\left(16\right)=k\cdot 4\left[{(\alpha +\beta )}^2-3\alpha \beta \right]\end{array}$
$\Rightarrow 4k=16-3k\Rightarrow k=\frac{16}{7}$