Question

The roots of $k{x}^3-26x^2+52x-24=0$ are in G.P, then $k$ is

• A. $3$
• B. $-3$
• C. $4$
• D. $-4$

Answer 1

The correct option is A $3$

As roots are in G.P then let $\alpha ,\beta ,\gamma$ are roots of $k{x}^3-26x^2+52x-24=0$
such that $\alpha \gamma ={\beta }^2$
$S_3=\alpha \beta \gamma =\frac{24}{k}\Rightarrow {\beta }^3=\frac{24}{k}\Rightarrow \beta ={\left(\frac{24}{k}\right)}^{\frac{1}{3}}$
Substituting in equation we get
$k{\left({\left(\frac{24}{k}\right)}^{\frac{1}{3}}\right)}^3-26{\left({\left(\frac{24}{k}\right)}^{\frac{1}{3}}\right)}^2+52{\left(\frac{24}{k}\right)}^{\frac{1}{3}}-24=0\Rightarrow -26{\left(\frac{24}{k}\right)}^{\frac{2}{3}}+52\left({\left(\frac{24}{k}\right)}^{\frac{1}{3}}\right)=0\Rightarrow k=3$