If the values of k for which the equation x^2 | vedclass

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(C) For the quadratic equation $x^2+2(k+2)x+6k+7=0$ to have equal roots,the discriminant must be zero,i.e.,$D = b^2 - 4ac = 0$. Comparing with $ax^2+b

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$10$

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(C) For the quadratic equation $x^2+2(k+2)x+6k+7=0$ to have equal roots,the discriminant must be zero,i.e.,$D = b^2 - 4ac = 0$. Comparing with $ax^2+bx+c=0$,we have $a=1$,$b=2(k+2)$,and $c=6k+7$. Substituting these values: $[2(k+2)]^2 - 4(1)(6k+7) = 0$ $4(k^2+4k+4) - 24k - 28 = 0$ $4k^2 + 16k + 16 - 24k - 28 = 0$ $4k^2 - 8k - 12 = 0$ Dividing by $4$: $k^2 - 2k - 3 = 0$ $(k-3)(k+1) = 0$ Thus,$k_1 = 3$ and $k_2 = -1$. Therefore,$k_1^2 + k_2^2 = (3)^2 + (-1)^2 = 9 + 1 = 10$.

If the values of $k$ for which the equation $x^2+2(k+2)x+6k+7=0$ has equal roots are $k_1$ and $k_2$,then $k_1^2+k_2^2=$

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