Sure, let's solve the inequality [tex]\( \frac{x^2 - 1}{x^2 + 4x + 1} < 1 \)[/tex] step by step:
1. Rewrite the Inequality:
First, subtract 1 from both sides of the inequality to set it to 0.
[tex]\[
\frac{x^2 - 1}{x^2 + 4x + 1} - 1 < 0
\][/tex]
2. Combine the Fractions:
Express the inequality as a single fraction.
[tex]\[
\frac{x^2 - 1}{x^2 + 4x + 1} - \frac{x^2 + 4x + 1}{x^2 + 4x + 1} < 0
\][/tex]
Simplify the numerator:
[tex]\[
\frac{x^2 - 1 - (x^2 + 4x + 1)}{x^2 + 4x + 1} < 0
\][/tex]
[tex]\[
\frac{x^2 - 1 - x^2 - 4x - 1}{x^2 + 4x + 1} < 0
\][/tex]
[tex]\[
\frac{-4x - 2}{x^2 + 4x + 1} < 0
\][/tex]
3. Factor the Numerator and Denominator:
Note that the numerator [tex]\(-4x - 2\)[/tex] can be factored as:
[tex]\[
-2(2x + 1)
\][/tex]
Therefore, our inequality is:
[tex]\[
\frac{-2(2x + 1)}{x^2 + 4x + 1} < 0
\][/tex]
Now we need to find the intervals where this fraction is negative.
4. Determine Critical Points:
Critical points occur where the numerator or denominator are zero. Set the numerator [tex]\(-2(2x + 1)\)[/tex] to zero:
[tex]\[
-2(2x + 1) = 0 \quad \Rightarrow \quad 2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}
\][/tex]
Set the denominator [tex]\(x^2 + 4x + 1\)[/tex] to zero. Find the roots using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[
x = \frac{-4 \pm \sqrt{16 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-4 \pm \sqrt{12}}{2}
\][/tex]
[tex]\[
x = \frac{-4 \pm 2\sqrt{3}}{2}
\][/tex]
[tex]\[
x = -2 \pm \sqrt{3}
\][/tex]
So, the roots are [tex]\( x = -2 - \sqrt{3} \)[/tex] and [tex]\( x = -2 + \sqrt{3} \)[/tex].
5. Test Intervals Between Critical Points:
Determine the sign of the fraction in the intervals defined by the critical points [tex]\( x = -2 - \sqrt{3} \)[/tex], [tex]\( x = -\frac{1}{2} \)[/tex], and [tex]\( x = -2 + \sqrt{3} \)[/tex].
- For [tex]\( x < -2 - \sqrt{3} \)[/tex]: Choose [tex]\( x = -4 \)[/tex]
- For [tex]\( -2 - \sqrt{3} < x < -\frac{1}{2} \)[/tex]: Choose [tex]\( x = -2 \)[/tex]
- For [tex]\( -\frac{1}{2} < x < -2 + \sqrt{3} \)[/tex]: Choose [tex]\( x = 0 \)[/tex]
- For [tex]\( x > -2 + \sqrt{3} \)[/tex]: Choose [tex]\( x = 2 \)[/tex]
Evaluate the sign of the fraction in these intervals. We find that the fraction is negative in the intervals
[tex]\[
(-2 - \sqrt{3}, -\frac{1}{2}) \cup (-2 + \sqrt{3}, \infty)
\][/tex]
6. Solution:
So, the solution to the inequality [tex]\( \frac{x^2 - 1}{x^2 + 4x + 1} < 1 \)[/tex] is
[tex]\[
\boxed{(-2 - \sqrt{3}, -\frac{1}{2}) \cup (-2 + \sqrt{3}, \infty)}
\][/tex]
