To determine the smallest integer value of [tex]\( q \)[/tex] for which the quadratic polynomial [tex]\( -x^2 + q x + (3 q + 8) \)[/tex] is always negative for all real values of [tex]\( x \)[/tex], we need to ensure that the polynomial never attains a positive or zero value, i.e., it remains strictly negative for every [tex]\( x \)[/tex].
A quadratic polynomial [tex]\( ax^2 + bx + c \)[/tex] remains always negative if:
1. The leading coefficient [tex]\( a \)[/tex] is negative.
2. The quadratic polynomial does not have any real roots (i.e., the discriminant [tex]\( \Delta \)[/tex] must be less than zero).
In our case:
[tex]\[ f(x) = -x^2 + qx + (3q + 8) \][/tex]
The leading coefficient [tex]\( a \)[/tex] is [tex]\(-1\)[/tex], which is already negative. Now, we need to check that the polynomial has no real roots.
The discriminant [tex]\( \Delta \)[/tex] of a quadratic polynomial [tex]\( ax^2 + bx + c \)[/tex] is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For [tex]\( f(x) = -x^2 + qx + (3q + 8) \)[/tex], we have:
[tex]\[ a = -1, \quad b = q, \quad c = 3q + 8 \][/tex]
Now, evaluate the discriminant:
[tex]\[ \Delta = q^2 - 4(-1)(3q + 8) \][/tex]
[tex]\[ \Delta = q^2 + 4(3q + 8) \][/tex]
[tex]\[ \Delta = q^2 + 12q + 32 \][/tex]
For the quadratic [tex]\( f(x) = -x^2 + qx + (3q + 8) \)[/tex] to be always negative, we need:
[tex]\[ \Delta < 0 \][/tex]
[tex]\[ q^2 + 12q + 32 < 0 \][/tex]
This is a quadratic inequality. To solve it, we first find the roots of the equation [tex]\( q^2 + 12q + 32 = 0 \)[/tex]:
[tex]\[ q = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \][/tex]
[tex]\[ q = \frac{-12 \pm \sqrt{144 - 128}}{2} \][/tex]
[tex]\[ q = \frac{-12 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ q = \frac{-12 \pm 4}{2} \][/tex]
[tex]\[ q = \frac{-12 + 4}{2} \quad \text{or} \quad q = \frac{-12 - 4}{2} \][/tex]
[tex]\[ q = \frac{-8}{2} \quad \text{or} \quad q = \frac{-16}{2} \][/tex]
[tex]\[ q = -4 \quad \text{or} \quad q = -8 \][/tex]
The roots of the quadratic are [tex]\( q = -4 \)[/tex] and [tex]\( q = -8 \)[/tex]. The quadratic [tex]\( q^2 + 12q + 32 \)[/tex] opens upwards (positive leading coefficient). Therefore, the inequality [tex]\( q^2 + 12q + 32 < 0 \)[/tex] is satisfied between the roots [tex]\( q = -8 \)[/tex] and [tex]\( q = -4 \)[/tex].
Hence, the values of [tex]\( q \)[/tex] that satisfy the inequality are:
[tex]\[ -8 < q < -4 \][/tex]
The smallest integer value of [tex]\( q \)[/tex] within this interval is:
[tex]\[ q = -7 \][/tex]
Thus, the smallest integer value of [tex]\( q \)[/tex] for which [tex]\( -x^2 + q x + (3 q + 8) \)[/tex] is always negative for all real values of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{-7} \][/tex]
