Question

If the roots of $x^3-7x^2+14x+k=0$ are in G.P. then k =

• A. $-8$
• B. $8$
• C. $7$
• D. $-7$

Answer 1

The correct option is A $-8$

Let $\alpha ,\beta ,\gamma$ be the roots of $x^3-7x^2+14x+k=0$
$\alpha +\beta +\gamma =7\alpha \beta +\beta \gamma +\gamma \alpha =14\alpha \beta \gamma =-k$
Now as $\alpha ,\beta ,\gamma$ is in G.P ${\beta }^2=\alpha \gamma$
$\alpha \beta \gamma =-k\Rightarrow {\beta }^3=-k\beta ={(-k)}^{\frac{1}{3}}$

Substituting this in equation we get
${\left({(-k)}^{\frac{1}{3}}\right)}^3-7{\left({(-k)}^{\frac{1}{2}}\right)}^3+14\left({(-k)}^{\frac{1}{3}}\right)+k=0\Rightarrow k=-8$