Determine k such that the quadratic equation | vedclass

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(A) Comparing the given equation $x^{2}-2(1+3k)x+7(3+2k)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a=1, b=-2(1+3k), c=7(3+2k)$For equal roots,

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

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(A) Comparing the given equation $x^{2}-2(1+3k)x+7(3+2k)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a=1, b=-2(1+3k), c=7(3+2k)$For equal roots,the discriminant $D$ must be $0$,i.e.,$D = b^{2}-4ac = 0$.Substituting the values:$[-2(1+3k)]^{2} - 4(1)(7(3+2k)) = 0$$4(1+3k)^{2} - 28(3+2k) = 0$Dividing by $4$:$(1+3k)^{2} - 7(3+2k) = 0$$1 + 9k^{2} + 6k - 21 - 14k = 0$$9k^{2} - 8k - 20 = 0$Using the quadratic formula $k = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$:$k = \frac{8 \pm \sqrt{(-8)^{2} - 4(9)(-20)}}{2(9)}$$k = \frac{8 \pm \sqrt{64 + 720}}{18} = \frac{8 \pm \sqrt{784}}{18}$$k = \frac{8 \pm 28}{18}$$k = \frac{36}{18} = 2$ or $k = \frac{-20}{18} = \frac{-10}{9}$.

Determine $k$ such that the quadratic equation $x^{2}-2(1+3k)x+7(3+2k)=0$ has equal roots.

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