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Question
The roots of the following quadratic equation is real and equal, find k.
3y2 + ky +12 = 0
Solution
3y2 + ky +12 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
\[b^2 - 4ac = 0\]
\[ \Rightarrow k^2 - 4 \times 3 \times 12 = 0\]
\[ \Rightarrow k^2 - 144 = 0\]
\[ \Rightarrow k^2 = 144\]
\[ \Rightarrow k = \pm 12\]
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