TS EAMCET · Maths · Quadratic Equation
If the quadratic equation \(3 x^2+(2 k+1) x-5 k=0\) has real and equal roots, then the value of \(k\) such that \(-\frac{1}{2} \lt k \lt 0\) is
- A \(\frac{-16+\sqrt{255}}{2}\)
- B \(\frac{-16-\sqrt{255}}{2}\)
- C \(-\frac{2}{3}\)
- D \(-\frac{3}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{-16+\sqrt{255}}{2}\)
Step-by-step Solution
Detailed explanation
\(D=0\) \(\begin{aligned} & (2 k+1)^2+4 \times 5 \times 3 k=0 \Rightarrow 4 k^2+64 k+1=0 \\ & \Rightarrow k=\frac{-64 \pm \sqrt{64^2-16}}{8}=\frac{-64 \pm 4 \sqrt{255}}{8} \\ & \Rightarrow k=\frac{-16+\sqrt{255}}{2} \in\left(\frac{-1}{2}, 0\right) .\end{aligned}\)
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