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If the roots of the equation $4 x^3-12 x^2+11 x+k=0$ are in arithmetic progression, then $k$ is equal to
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1812 Upvotes
Verified Answer
The correct answer is:
$-3$
Since, the roots of the equation $4 x^3-12 x^2$ $+11 x+k=0$ are in A.P is $\alpha-a, \alpha, \alpha+a$
$\therefore$ Sum of roots, $\quad 3 \alpha=\frac{12}{4}=3$
$$
\Rightarrow \quad \alpha=1
$$
Since, $\alpha$ is a root, therefore it satisfies the given equation
$$
\Rightarrow \quad 4-12+11+k=0 \Rightarrow k=-3 \text {. }
$$
$\therefore$ Sum of roots, $\quad 3 \alpha=\frac{12}{4}=3$
$$
\Rightarrow \quad \alpha=1
$$
Since, $\alpha$ is a root, therefore it satisfies the given equation
$$
\Rightarrow \quad 4-12+11+k=0 \Rightarrow k=-3 \text {. }
$$
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