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If the roots of $x^3-k x^2+14 x-8=0$ are in geometric progression, then $k$ is equal to
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The correct answer is:
$7$
Given equation, $x^3-k x^2+14 x-8=0$
Roots of the above equation is in G.P. So let $\frac{a}{r}, a$ and $a r$ are the roots of equation.
Then, product of roots
$\begin{aligned}
& \frac{a}{r} \cdot a \cdot a r=\frac{D}{A} \Rightarrow \frac{a}{r} \cdot a \cdot a r=8 \Rightarrow a^3=8 \\
& \therefore \quad a=2
\end{aligned}$
Therefore, at $a=2$ is the root of equation.
$\begin{aligned}
& \therefore \quad x^3-k x^2+14 x-8=0 \\
& (2)^3-k(2)^2+14(2)-8=0 \\
& 8-4 k+28-8=0 \Rightarrow k=7
\end{aligned}$
Roots of the above equation is in G.P. So let $\frac{a}{r}, a$ and $a r$ are the roots of equation.
Then, product of roots
$\begin{aligned}
& \frac{a}{r} \cdot a \cdot a r=\frac{D}{A} \Rightarrow \frac{a}{r} \cdot a \cdot a r=8 \Rightarrow a^3=8 \\
& \therefore \quad a=2
\end{aligned}$
Therefore, at $a=2$ is the root of equation.
$\begin{aligned}
& \therefore \quad x^3-k x^2+14 x-8=0 \\
& (2)^3-k(2)^2+14(2)-8=0 \\
& 8-4 k+28-8=0 \Rightarrow k=7
\end{aligned}$
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