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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

  1. A \(\frac{-16+\sqrt{255}}{2}\)
  2. B \(\frac{-16-\sqrt{255}}{2}\)
  3. C \(-\frac{2}{3}\)
  4. D \(-\frac{3}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{-16+\sqrt{255}}{2}\)

Step-by-step Solution

Detailed explanation

\(D=0\) \(\begin{aligned} & (2 k+1)^2+4 \times 5 \times 3 k=0 \Rightarrow 4 k^2+64 k+1=0 \\ & \Rightarrow k=\frac{-64 \pm \sqrt{64^2-16}}{8}=\frac{-64 \pm 4 \sqrt{255}}{8} \\ & \Rightarrow k=\frac{-16+\sqrt{255}}{2} \in\left(\frac{-1}{2}, 0\right) .\end{aligned}\)