Question

e solution for $x^2 + 36 > 12x$
6}
< 6}
$\mathbb{r}$ }
$\mathbb{r}$ and $x \
eq 6$
the solution for $x^2 + 2x + 8 \leq 0$ is
all real numbers
the empty set
x = 2 or x = 4
x = -2 or x = 4

Explanation:

Step1: Analyze the quadratic inequality

We have the inequality \(x^{2}+2x + 8\leqslant0\). Let's consider the quadratic function \(y=x^{2}+2x + 8\). The discriminant of a quadratic function \(ax^{2}+bx + c\) is given by \(\Delta=b^{2}-4ac\). For \(y=x^{2}+2x + 8\), \(a = 1\), \(b=2\) and \(c = 8\). So \(\Delta=(2)^{2}-4\times1\times8=4 - 32=- 28\).

Step2: Determine the nature of the quadratic function

Since \(a = 1>0\), the parabola \(y=x^{2}+2x + 8\) opens upwards. And since \(\Delta=-28<0\), the quadratic function \(y=x^{2}+2x + 8\) has no real roots and is always positive for all real values of \(x\) (because the parabola opens upwards and never intersects the \(x\)-axis). So the inequality \(x^{2}+2x + 8\leqslant0\) has no real solutions.

Answer:

the empty set