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(C) Comparing the equation $(k+1) x^{2}-2(k+3) x+(2 k+3)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a = k+1, b = -2(k+3), c = 2k+3$Since the qu

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$3, -2$

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(C) Comparing the equation $(k+1) x^{2}-2(k+3) x+(2 k+3)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a = k+1, b = -2(k+3), c = 2k+3$Since the quadratic equation has two equal and real roots,the discriminant $D$ must be zero.$D = b^{2} - 4ac = 0$$(-2(k+3))^{2} - 4(k+1)(2k+3) = 0$$4(k^{2} + 6k + 9) - 4(2k^{2} + 3k + 2k + 3) = 0$Divide by $4$:$(k^{2} + 6k + 9) - (2k^{2} + 5k + 3) = 0$$k^{2} + 6k + 9 - 2k^{2} - 5k - 3 = 0$$-k^{2} + k + 6 = 0$$k^{2} - k - 6 = 0$$(k-3)(k+2) = 0$Therefore,$k = 3$ or $k = -2$.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $(k+1) x^{2}-2(k+3) x+(2 k+3)=0$

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