To solve the inequality [tex]\((x-4)(x+2) > 7\)[/tex], we need to proceed by manipulating it into a form that we can analyze more conveniently. Here are the steps:
1. Set up the inequality:
[tex]\[
(x-4)(x+2) > 7
\][/tex]
2. Isolate the quadratic expression:
Subtract 7 from both sides to focus on finding the critical points where the expression equals zero:
[tex]\[
(x-4)(x+2) - 7 > 0
\][/tex]
3. Factor the quadratic expression:
Let's set the expression equal to zero to find the critical points where changes might occur in the inequality:
[tex]\[
(x-4)(x+2) - 7 = 0
\][/tex]
This can be expanded to:
[tex]\[
x^2 - 4x + 2x - 8 = 7
\][/tex]
Combine like terms:
[tex]\[
x^2 - 2x - 8 = 7
\][/tex]
Then move 7 to the left side:
[tex]\[
x^2 - 2x - 15 = 0
\][/tex]
4. Solve the quadratic equation:
Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a=1\)[/tex], [tex]\(b=-2\)[/tex], and [tex]\(c=-15\)[/tex]:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{4 + 60}}{2}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{64}}{2}
\][/tex]
[tex]\[
x = \frac{2 \pm 8}{2}
\][/tex]
Hence, the solutions are:
[tex]\[
x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-6}{2} = -3
\][/tex]
5. Determine the intervals for testing:
The quadratic equation [tex]\(x^2 - 2x - 15 = 0\)[/tex] has the roots [tex]\(x = -3\)[/tex] and [tex]\(x = 5\)[/tex]. These points divide the number line into three intervals: [tex]\((- \infty, -3)\)[/tex], [tex]\((-3, 5)\)[/tex], and [tex]\((5, \infty)\)[/tex].
6. Test the intervals:
To find where the inequality [tex]\((x-4)(x+2) > 7\)[/tex] holds true, we test points within these intervals:
- For the interval [tex]\((- \infty, -3)\)[/tex], pick [tex]\(x = -4\)[/tex]:
[tex]\[
(x-4)(x+2) - 7 = (-4-4)(-4+2) - 7 = (-8)(-2) - 7 = 16 - 7 = 9 > 0
\][/tex]
This is true, so the inequality holds in [tex]\((- \infty, -3)\)[/tex].
- For the interval [tex]\((-3, 5)\)[/tex], pick [tex]\(x = 0\)[/tex]:
[tex]\[
(x-4)(x+2) - 7 = (0-4)(0+2) - 7 = (-4)(2) - 7 = -8 - 7 = -15 < 0
\][/tex]
This is false, so the inequality does not hold in [tex]\((-3, 5)\)[/tex].
- For the interval [tex]\((5, \infty)\)[/tex], pick [tex]\(x = 6\)[/tex]:
[tex]\[
(x-4)(x+2) - 7 = (6-4)(6+2) - 7 = (2)(8) - 7 = 16 - 7 = 9 > 0
\][/tex]
This is true, so the inequality holds in [tex]\((5, \infty)\)[/tex].
Therefore, the solution for [tex]\((x-4)(x+2) > 7\)[/tex] is:
[tex]\[
(-\infty < x < -3) \cup (5 < x < \infty)
\][/tex]
