To solve the inequality [tex]\(\frac{2x^2 - 3x - 459}{x^2 + 1} > 1\)[/tex], let's work through it step-by-step.
1. Rewrite the Inequality:
The given inequality is:
[tex]\[
\frac{2x^2 - 3x - 459}{x^2 + 1} > 1
\][/tex]
2. Subtract 1 from Both Sides:
To simplify the inequality, we can subtract 1 from both sides:
[tex]\[
\frac{2x^2 - 3x - 459}{x^2 + 1} - 1 > 0
\][/tex]
3. Combine Like Terms:
Rewrite the left-hand side by combining the fractions:
[tex]\[
\frac{2x^2 - 3x - 459 - (x^2 + 1)}{x^2 + 1} > 0
\][/tex]
Simplifying the numerator:
[tex]\[
\frac{2x^2 - 3x - 459 - x^2 - 1}{x^2 + 1} > 0
\][/tex]
[tex]\[
\frac{x^2 - 3x - 460}{x^2 + 1} > 0
\][/tex]
4. Find the Roots of the Numerator:
To understand where the rational function changes its sign, we need to find the roots of the numerator [tex]\(x^2 - 3x - 460\)[/tex]. We solve the quadratic equation:
[tex]\[
x^2 - 3x - 460 = 0
\][/tex]
The roots can be found using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-460)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{9 + 1840}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm \sqrt{1849}}{2}
\][/tex]
[tex]\[
x = \frac{3 \pm 43}{2}
\][/tex]
So, the roots are:
[tex]\[
x = \frac{3 + 43}{2} = 23 \quad \text{and} \quad x = \frac{3 - 43}{2} = -20
\][/tex]
5. Analyze the Intervals:
Now, we need to analyze the sign of [tex]\(\frac{x^2 - 3x - 460}{x^2 + 1}\)[/tex] over the intervals determined by the roots [tex]\(x = -20\)[/tex] and [tex]\(x = 23\)[/tex]. These intervals are:
[tex]\[
(-\infty, -20), (-20, 23), \text{and} (23, \infty)
\][/tex]
6. Test Points in Each Interval:
For each interval, we test a point to determine the sign of the expression:
- For [tex]\(x \in (-\infty, -20)\)[/tex], choose [tex]\(x = -21\)[/tex]:
[tex]\[
\frac{(-21)^2 - 3(-21) - 460}{(-21)^2 + 1} = \frac{441 + 63 - 460}{441 + 1} > 0
\][/tex]
- For [tex]\(x \in (-20, 23)\)[/tex], choose [tex]\(x = 0\)[/tex]:
[tex]\[
\frac{0^2 - 3\cdot0 - 460}{0^2 + 1} = \frac{-460}{1} < 0
\][/tex]
- For [tex]\(x \in (23, \infty)\)[/tex], choose [tex]\(x = 24\)[/tex]:
[tex]\[
\frac{24^2 - 3\cdot24 - 460}{24^2 + 1} = \frac{576 - 72 - 460}{576 + 1} > 0
\][/tex]
7. Combine the Results:
The expression is positive in the intervals [tex]\((- \infty, -20)\)[/tex] and [tex]\((23, \infty)\)[/tex].
Therefore, the solution to the inequality [tex]\(\frac{2x^2 - 3x - 459}{x^2 + 1} > 1\)[/tex] is:
[tex]\[
x \in (-\infty, -20) \cup (23, \infty)
\][/tex]
