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

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Question

(C) For the quadratic equation ${x^2} - 8x + ({a^2} - 6a) = 0$ to have real roots,the discriminant $D$ must be greater than or equal to zero $(D \ge 0

Vedclass · Accepted Answer

$ - 2 \le a \le 8$

Vedclass · Answer

(C) For the quadratic equation ${x^2} - 8x + ({a^2} - 6a) = 0$ to have real roots,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.The discriminant $D = b^2 - 4ac$,where $a = 1$,$b = -8$,and $c = ({a^2} - 6a)$.Substituting these values: $D = (-8)^2 - 4(1)({a^2} - 6a) \ge 0$.$64 - 4({a^2} - 6a) \ge 0$.Dividing by $4$: $16 - ({a^2} - 6a) \ge 0$.$16 - {a^2} + 6a \ge 0$.Multiplying by $-1$ (reversing the inequality sign): ${a^2} - 6a - 16 \le 0$.Factoring the quadratic expression: $(a - 8)(a + 2) \le 0$.For the product of two factors to be less than or equal to zero,the value of $a$ must lie between the roots of the equation $(a - 8)(a + 2) = 0$,which are $a = 8$ and $a = -2$.Thus,the range of $a$ is $-2 \le a \le 8$.

If the roots of the equation ${x^2} - 8x + ({a^2} - 6a) = 0$ are real,then

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