If the roots of the equation $x^2 - 8x + a^2 - 6a = 0$ are real, then the value of $a$ will be
- $-2 < a < 8$
- $-2 \le a \le 8$
- $2 < a < 8$
- $2 \le a \le 8$
For real roots, the discriminant D >= 0. D = (-8)^2 - 4(1)(a^2 - 6a) = 64 - 4a^2 + 24a >= 0. Dividing by -4: a^2 - 6a - 16 <= 0. Factors: (a-8)(a+2) <= 0. Thus, -2 <= a <= 8.
For the roots of the quadratic equation x squared minus 8x plus a squared minus 6a to be real, the discriminant must be greater than or equal to zero. The discriminant is (negative 8) squared minus 4 times 1 times (a squared minus 6a), which simplifies to 64 minus 4a squared plus 24a. Dividing by negative 4 and flipping the inequality gives a squared minus 6a minus 16 being less than or equal to 0. Factoring this yields (a minus 8)(a plus 2) is less than or equal to 0, which means a lies between negative 2 and 8 inclusive.