If the roots of the equation $x^{2} + a^{2} = 8x + 6a$ are real, then:
- $a \epsilon \left [ 2,8 \right ]$
- $ a \epsilon \left [ -2,8 \right ]$
- $a \epsilon (2,8)$
- $ a \epsilon (-2,8)$
For real roots, the discriminant D >= 0. Equation: x^2 - 8x + (a^2 - 6a) = 0. D = (-8)^2 - 4(1)(a^2 - 6a) = 64 - 4a^2 + 24a. 64 + 24a - 4a^2 >= 0 implies a^2 - 6a - 16 <= 0. Factoring: (a - 8)(a + 2) <= 0. This holds for a in [-2, 8].
Rearrange the given equation into the standard quadratic form: x^2 - 8x + (a^2 - 6a) = 0. For the roots to be real, the discriminant must be greater than or equal to zero, meaning (-8)^2 - 4(1)(a^2 - 6a) >= 0. This simplifies to 64 - 4a^2 + 24a >= 0, and dividing by -4 gives a^2 - 6a - 16 <= 0. Factizing this quadratic inequality yields (a - 8)(a + 2) <= 0, which holds true when a lies between the roots: a belongs to the closed interval [-2, 8].