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If the roots of the equation $3 x^2+4 k x+3=0$ are non-real, then $\mathrm{k}$ lies in the interval
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Verified Answer
The correct answer is:
$\left[\frac{-3}{2}, \frac{3}{2}\right]$
Given: $3 \mathrm{x}^2+4 \mathrm{kx}+3=0$
Above eq ${ }^{\mathrm{n}}$ has non-real roots if
$$
\begin{aligned}
& (4 \mathrm{k})^2-4 \times 3 \times 3 < 0 \\
& \Rightarrow 16 \mathrm{k}^2-36 < 0 \\
& \Rightarrow \mathrm{k}^2 < \frac{9}{4} \\
& \Rightarrow-\frac{3}{2} < \mathrm{k} < \frac{3}{2}
\end{aligned}
$$
$\Rightarrow \mathrm{k}$ lies in the interval $\left(-\frac{3}{2}, \frac{3}{2}\right)$
Above eq ${ }^{\mathrm{n}}$ has non-real roots if
$$
\begin{aligned}
& (4 \mathrm{k})^2-4 \times 3 \times 3 < 0 \\
& \Rightarrow 16 \mathrm{k}^2-36 < 0 \\
& \Rightarrow \mathrm{k}^2 < \frac{9}{4} \\
& \Rightarrow-\frac{3}{2} < \mathrm{k} < \frac{3}{2}
\end{aligned}
$$
$\Rightarrow \mathrm{k}$ lies in the interval $\left(-\frac{3}{2}, \frac{3}{2}\right)$
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