Question

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

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

Answer 1

Answer: (B)

Given $\alpha \beta ; \alpha \beta(\alpha+\beta) ; \alpha^3+\beta^3$ are in G.P.
$\alpha+\beta=4 ; \alpha \beta= k ; \alpha \beta^2+\alpha^2 \beta=\alpha \beta(\alpha+\beta)=4 k $
$\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta) $
$=64-3 k (4)=4(16-3 k ) $
$\therefore k ; 4 k ; 4(16-3 k ) \text { are in G.P. } $
$16 k ^2=4 k (16-3 k ) $
$4 k (4 k -16+3 k )=0$
$k =0 ; k =\frac{16}{7} $