If the quadratic equation 3x^2 + (2k + 1)x - | vedclass

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(A) For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Given equation: $3

Vedclass · Accepted Answer

$\frac{-16 + \sqrt{255}}{2}$

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(A) For a quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Given equation: $3x^2 + (2k + 1)x - 5k = 0$.Here,$a = 3$,$b = (2k + 1)$,and $c = -5k$.$D = (2k + 1)^2 - 4(3)(-5k) = 0$.$4k^2 + 4k + 1 + 60k = 0$.$4k^2 + 64k + 1 = 0$.Using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$k = \frac{-64 \pm \sqrt{64^2 - 4(4)(1)}}{2(4)} = \frac{-64 \pm \sqrt{4096 - 16}}{8} = \frac{-64 \pm \sqrt{4080}}{8}$.$k = \frac{-64 \pm 4\sqrt{255}}{8} = \frac{-16 \pm \sqrt{255}}{2}$.Since $-\frac{1}{2} $k_1 = \frac{-16 + 15.96}{2} \approx -0.02$ (which is in the range $(-\frac{1}{2}, 0)$).$k_2 = \frac{-16 - 15.96}{2} \approx -15.98$ (which is not in the range).

If the quadratic equation $3x^2 + (2k + 1)x - 5k = 0$ has real and equal roots,then the value of $k$ such that $-\frac{1}{2} < k < 0$ is

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