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(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = 1$,$b = -2(1 + 3k

Vedclass · Accepted Answer

$2, - \frac{10}{9}$

Vedclass · Answer

(B) For a quadratic equation $ax^2 + bx + c = 0$ to have equal roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = 1$,$b = -2(1 + 3k)$,and $c = 7(3 + 2k)$.Setting $D = 0$:$[-2(1 + 3k)]^2 - 4(1)(7(3 + 2k)) = 0$$4(1 + 3k)^2 - 28(3 + 2k) = 0$Divide by $4$:$(1 + 3k)^2 - 7(3 + 2k) = 0$$1 + 6k + 9k^2 - 21 - 14k = 0$$9k^2 - 8k - 20 = 0$Solving the quadratic equation $9k^2 - 8k - 20 = 0$ using the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$k = \frac{8 \pm \sqrt{(-8)^2 - 4(9)(-20)}}{2(9)}$$k = \frac{8 \pm \sqrt{64 + 720}}{18}$$k = \frac{8 \pm \sqrt{784}}{18}$$k = \frac{8 \pm 28}{18}$$k_1 = \frac{36}{18} = 2$ and $k_2 = \frac{-20}{18} = -\frac{10}{9}$.Thus,the values of $k$ are $2, -\frac{10}{9}$.

For what values of $k$ will the equation $x^2 - 2(1 + 3k)x + 7(3 + 2k) = 0$ have equal roots?

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