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

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Question

(C) The roots of the quadratic equation $x^2 - 8x + (a^2 - 6a) = 0$ are real if and only if the discriminant $D \ge 0$.For the equation $Ax^2 + Bx + C

Vedclass · Accepted Answer

$ - 2 \le a \le 8$

Vedclass · Answer

(C) The roots of the quadratic equation $x^2 - 8x + (a^2 - 6a) = 0$ are real if and only if the discriminant $D \ge 0$.For the equation $Ax^2 + Bx + C = 0$,$D = B^2 - 4AC$.Here,$A = 1$,$B = -8$,and $C = a^2 - 6a$.$D = (-8)^2 - 4(1)(a^2 - 6a) \ge 0$$64 - 4(a^2 - 6a) \ge 0$$16 - (a^2 - 6a) \ge 0$$16 - a^2 + 6a \ge 0$$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 inequality $(a - 8)(a + 2) \le 0$ holds for $a \in [-2, 8]$.Therefore,$-2 \le a \le 8$.

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

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