If the roots of the equation x^2 + a^2 = 8x + | vedclass

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(B) The given equation is $x^2 - 8x + (a^2 - 6a) = 0$.For the roots to be real,the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4a

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$a \in [-2, 8]$

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(B) The given equation is $x^2 - 8x + (a^2 - 6a) = 0$.For the roots to be real,the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac \ge 0$$(-8)^2 - 4(1)(a^2 - 6a) \ge 0$$64 - 4a^2 + 24a \ge 0$Dividing by $-4$ (and reversing the inequality sign):$a^2 - 6a - 16 \le 0$Factoring the quadratic:$(a - 8)(a + 2) \le 0$This inequality holds when $a$ lies between the roots $-2$ and $8$,inclusive.Therefore,$a \in [-2, 8]$.

If the roots of the equation $x^2 + a^2 = 8x + 6a$ are real,then

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