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Two roots of the cubic $x^3 + 3x^2 + kx = 0$ the value of k is 12 = 0 are real and unequal but have the same absolute value. Then
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Verified Answer
The correct answer is:
$-4$
\begin{align*}
&\text{Let the roots be } r, -r, \text{ and } t. \text{ Hence, } r - r + t = -3 \Rightarrow t = -3. \\
&\text{Now, the product of the roots is } (r)(-r)(t) = 12. \\
&\Rightarrow -r^2 \cdot (-3) = 12. \\
&\Rightarrow r^2 = 4. \\
&\text{Hence, the roots are } 2, -2, \text{ and } -3. \\
&\text{Now, the sum of the product taken two at a time is } r(-r) + (-r)t + tr = k. \\
&\Rightarrow -4 - 6 + 6 = k. \\
&\Rightarrow k = -4.
\end{align*}
&\text{Let the roots be } r, -r, \text{ and } t. \text{ Hence, } r - r + t = -3 \Rightarrow t = -3. \\
&\text{Now, the product of the roots is } (r)(-r)(t) = 12. \\
&\Rightarrow -r^2 \cdot (-3) = 12. \\
&\Rightarrow r^2 = 4. \\
&\text{Hence, the roots are } 2, -2, \text{ and } -3. \\
&\text{Now, the sum of the product taken two at a time is } r(-r) + (-r)t + tr = k. \\
&\Rightarrow -4 - 6 + 6 = k. \\
&\Rightarrow k = -4.
\end{align*}
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