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What is the solution to [tex]$x^2 + 4x \ \textgreater \ 12$[/tex]?

A. [tex]x \ \textless \ -6[/tex] or [tex]x \ \textgreater \ 2[/tex]
B. [tex]-6 \ \textless \ x \ \textless \ 2[/tex]
C. [tex]x \ \textless \ -2[/tex] or [tex]x \ \textgreater \ 6[/tex]
D. [tex]-2 \ \textless \ x \ \textless \ 6[/tex]


Sagot :

To solve the inequality [tex]\(x^2 + 4x > 12\)[/tex], let's follow these steps:

1. Rewrite the inequality:
[tex]\[ x^2 + 4x - 12 > 0 \][/tex]

2. Find the roots of the corresponding equation:
[tex]\[ x^2 + 4x - 12 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -12\)[/tex].

Calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-12) = 16 + 48 = 64 \][/tex]

So the roots are:
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{2 \cdot 1} = \frac{-4 \pm 8}{2} \][/tex]

This yields two roots:
[tex]\[ x_1 = \frac{-4 + 8}{2} = 2 \quad \text{and} \quad x_2 = \frac{-4 - 8}{2} = -6 \][/tex]

3. Determine the intervals:
The roots divide the real number line into three intervals:
- [tex]\( (-\infty, -6) \)[/tex]
- [tex]\( (-6, 2) \)[/tex]
- [tex]\( (2, \infty) \)[/tex]

4. Test points from each interval to determine where the inequality is satisfied:
Choose a test point in each interval and substitute it into the inequality [tex]\(x^2 + 4x - 12 > 0\)[/tex]:

- For the interval [tex]\((-\infty, -6)\)[/tex], choose [tex]\(x = -7\)[/tex]:
[tex]\[ (-7)^2 + 4(-7) - 12 = 49 - 28 - 12 = 49 - 40 = 9 > 0 \][/tex]
This interval satisfies the inequality.

- For the interval [tex]\((-6, 2)\)[/tex], choose [tex]\(x = 0\)[/tex]:
[tex]\[ 0^2 + 4 \cdot 0 - 12 = -12 < 0 \][/tex]
This interval does not satisfy the inequality.

- For the interval [tex]\((2, \infty)\)[/tex], choose [tex]\(x = 3\)[/tex]:
[tex]\[ 3^2 + 4 \cdot 3 - 12 = 9 + 12 - 12 = 9 > 0 \][/tex]
This interval satisfies the inequality.

5. Combine the intervals where the inequality holds:
The solution is where the inequality is satisfied, which is [tex]\(x \in (-\infty, -6) \cup (2, \infty)\)[/tex].

Therefore, the correct solution to the inequality [tex]\(x^2 + 4x > 12\)[/tex] is:
[tex]\[ x < -6 \quad \text{or} \quad x > 2 \][/tex]

Hence, the answer is:
[tex]\[ \boxed{x < -6 \quad \text{or} \quad x > 2} \][/tex]