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(D) Comparing the given quadratic equation $x^{2}-(k+10)x+9(k+1)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a=1, b=-(k+10), c=9(k+1)$.Since the

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$8$

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(D) Comparing the given quadratic equation $x^{2}-(k+10)x+9(k+1)=0$ with the standard form $ax^{2}+bx+c=0$,we get:$a=1, b=-(k+10), c=9(k+1)$.Since the quadratic equation has two equal and real roots,the discriminant $D$ must be equal to $0$.$D = b^{2}-4ac = 0$Substituting the values:$( -(k+10) )^{2} - 4(1)(9(k+1)) = 0$$(k+10)^{2} - 36(k+1) = 0$$k^{2} + 20k + 100 - 36k - 36 = 0$$k^{2} - 16k + 64 = 0$This is a perfect square trinomial:$(k-8)^{2} = 0$Taking the square root on both sides:$k-8 = 0$$k = 8$Therefore,the value of $k$ is $8$.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $x^{2}-(k+10)x+9(k+1)=0$.

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