Q52 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

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

  • Correct answer
    A
    $\frac{-16+\sqrt{255}}{2}$
  • B
    $\frac{-16-\sqrt{255}}{2}$
  • C
    $-\frac{2}{3}$
  • D
    $-\frac{3}{5}$

Solution

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

Hence, the correct answer is option: 'A'.

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