If the roots of the equation x^2 - 8x + (a^2 | 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 $0$.$D = b^2 - 4ac

Vedclass · Accepted Answer

$-2 \leq a \leq 8$

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 $0$.$D = b^2 - 4ac \geq 0$$(-8)^2 - 4(1)(a^2 - 6a) \geq 0$$64 - 4(a^2 - 6a) \geq 0$Divide by $4$:$16 - (a^2 - 6a) \geq 0$$16 - a^2 + 6a \geq 0$$a^2 - 6a - 16 \leq 0$Factor the quadratic expression:$(a - 8)(a + 2) \leq 0$This inequality holds when $a$ lies between the roots of the equation $(a - 8)(a + 2) = 0$,which are $a = 8$ and $a = -2$.Thus,$-2 \leq a \leq 8$.

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

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