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

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$0$ and $3$

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(A) For a quadratic equation $ax^2 + bx + c = 0$ to have two equal and real roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = (k+1)$,$b = -2(k-1)$,and $c = 1$.Substituting these values into the discriminant formula:$D = [-2(k-1)]^2 - 4(k+1)(1) = 0$$4(k-1)^2 - 4(k+1) = 0$Dividing by $4$:$(k-1)^2 - (k+1) = 0$$k^2 - 2k + 1 - k - 1 = 0$$k^2 - 3k = 0$$k(k-3) = 0$Therefore,$k = 0$ or $k = 3$.

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

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