To determine the range of [tex]\(a\)[/tex] such that the roots of the quadratic equation [tex]\(x^2 + a^2 = 8x + 6a\)[/tex] are real, we need to rewrite the equation in the standard form and analyze its discriminant.
Starting with:
[tex]\[ x^2 + a^2 = 8x + 6a \][/tex]
First, rearrange it into the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ x^2 - 8x + a^2 - 6a = 0 \][/tex]
Here, the quadratic equation is:
[tex]\[ x^2 - 8x + (a^2 - 6a) = 0 \][/tex]
For the roots of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] to be real, the discriminant [tex]\(\Delta\)[/tex] must be greater than or equal to zero. The discriminant [tex]\(\Delta\)[/tex] for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -8\)[/tex], and [tex]\(c = a^2 - 6a\)[/tex]. So, the discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = (-8)^2 - 4 \cdot 1 \cdot (a^2 - 6a) \][/tex]
[tex]\[ \Delta = 64 - 4(a^2 - 6a) \][/tex]
[tex]\[ \Delta = 64 - 4a^2 + 24a \][/tex]
[tex]\[ \Delta = 64 + 24a - 4a^2 \][/tex]
For the roots to be real, we require:
[tex]\[ 64 + 24a - 4a^2 \geq 0 \][/tex]
This is a quadratic inequality in [tex]\(a\)[/tex]. Let's solve it by finding the roots of the associated quadratic equation:
[tex]\[ -4a^2 + 24a + 64 = 0 \][/tex]
Rewriting it in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ 4a^2 - 24a - 64 = 0 \][/tex]
Now, solve this quadratic equation using the quadratic formula [tex]\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 4\)[/tex], [tex]\(b = -24\)[/tex], and [tex]\(c = -64\)[/tex]:
[tex]\[ a = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 4 \cdot (-64)}}{2 \cdot 4} \][/tex]
[tex]\[ a = \frac{24 \pm \sqrt{576 + 1024}}{8} \][/tex]
[tex]\[ a = \frac{24 \pm \sqrt{1600}}{8} \][/tex]
[tex]\[ a = \frac{24 \pm 40}{8} \][/tex]
This gives us two solutions:
[tex]\[ a = \frac{24 + 40}{8} = \frac{64}{8} = 8 \][/tex]
[tex]\[ a = \frac{24 - 40}{8} = \frac{-16}{8} = -2 \][/tex]
The roots of the quadratic equation are [tex]\(a = 8\)[/tex] and [tex]\(a = -2\)[/tex]. The quadratic expression [tex]\(64 + 24a - 4a^2\)[/tex] is a downward facing parabola (since the coefficient of [tex]\(a^2\)[/tex] is negative). The expression is non-negative between the roots [tex]\(a = -2\)[/tex] and [tex]\(a = 8\)[/tex].
Thus, the interval for [tex]\(a\)[/tex] to ensure real roots is:
[tex]\[ a \in [-2, 8] \][/tex]
Therefore, the correct option is:
2) [tex]\([-2, 8]\)[/tex]
