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(A) Given the quadratic equation $f(x) = x^2 - 2kx + k^2 + k - 5 = 0$.For both roots to be less than $5$,three conditions must be satisfied:$1$. Discr

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$( - \infty , 4)$

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(A) Given the quadratic equation $f(x) = x^2 - 2kx + k^2 + k - 5 = 0$.For both roots to be less than $5$,three conditions must be satisfied:$1$. Discriminant $D \ge 0$: $D = (-2k)^2 - 4(1)(k^2 + k - 5) = 4k^2 - 4k^2 - 4k + 20 = 20 - 4k$. Setting $20 - 4k \ge 0$,we get $k \le 5$.$2$. Vertex position: The $x$-coordinate of the vertex $-b/(2a) = 2k/2 = k$ must be less than $5$,so $k $3$. Function value at $x = 5$: Since the parabola opens upward,$f(5) > 0$. Thus,$5^2 - 2k(5) + k^2 + k - 5 > 0$,which simplifies to $25 - 10k + k^2 + k - 5 > 0$,or $k^2 - 9k + 20 > 0$. Factoring gives $(k - 4)(k - 5) > 0$,which implies $k  5$.Combining all conditions: $(k \le 5) \cap (k  5)$,we get $k < 4$. Thus,$k \in (- \infty, 4)$.

If both the roots of the quadratic equation ${x^2} - 2kx + {k^2} + k - 5 = 0$ are less than $5$,then $k$ lies in the interval

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