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

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(B) For the quadratic equation $Ax^2 + Bx + C = 0$ to have real roots,the discriminant $D = B^2 - 4AC$ must be greater than or equal to $0$.Here,$A =

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$-2 \leq a \leq 8$

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(B) For the quadratic equation $Ax^2 + Bx + C = 0$ to have real roots,the discriminant $D = B^2 - 4AC$ must be greater than or equal to $0$.Here,$A = 1$,$B = -8$,and $C = a^2 - 6a$.Substituting these values into the discriminant formula:$D = (-8)^2 - 4(1)(a^2 - 6a) \geq 0$$64 - 4(a^2 - 6a) \geq 0$$64 - 4a^2 + 24a \geq 0$Dividing by $-4$ (and reversing the inequality sign):$a^2 - 6a - 16 \leq 0$Factoring the quadratic expression:$(a - 8)(a + 2) \leq 0$The roots of the inequality are $a = 8$ and $a = -2$.For the expression to be $\leq 0$,$a$ must lie between the roots inclusive:$-2 \leq a \leq 8$.

If the roots of the equation $x^2 - 8x + a^2 - 6a = 0$ are real,then what is the range of $a$?

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