Question

Find the number of real solutions for the equation x^3 + 2x^2 - 7x - 12 = 0

Answer 1

To determine the number of real solutions for the equation \( x^3 + 2x^2 - 7x - 12 = 0 \), we can use calculus to analyze its behavior.

1. **Find the first derivative**:
\[
f(x) = x^3 + 2x^2 - 7x - 12
\]
\[
f'(x) = 3x^2 + 4x - 7
\]

2. **Find the critical points** by setting the first derivative to zero:
\[
3x^2 + 4x - 7 = 0
\]
We can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 4 \), and \( c = -7 \).
\[
b^2 - 4ac = 4^2 - 4(3)(-7) = 16 + 84 = 100
\]
The critical points are:
\[
x = \frac{-4 \pm \sqrt{100}}{2 \cdot 3} = \frac{-4 \pm 10}{6}
\]
\[
x_1 = \frac{6}{6} = 1, \quad x_2 = \frac{-14}{6} = -\frac{7}{3}
\]

3. **Evaluate the function at the critical points**:
- Calculate \( f(-\frac{7}{3}) \):
\[
f(-\frac{7}{3}) = \left(-\frac{7}{3}\right)^3 + 2\left(-\frac{7}{3}\right)^2 - 7\left(-\frac{7}{3}\right) - 12
\]
\[
= -\frac{343}{27} + 2 \cdot \frac{49}{9} + \frac{49}{3} - 12
\]
\[
= -\frac{343}{27} + \frac{98}{9} + \frac{441}{27} - 12
\]
\[
= -\frac{343}{27} + \frac{294}{27} + \frac{441}{27} - \frac{324}{27}
\]
\[
= \frac{-343 + 294 + 441 - 324}{27} = \frac{68}{27} > 0
\]

- Calculate \( f(1) \):
\[
f(1) = 1^3 + 2 \cdot 1^2 - 7 \cdot 1 - 12 = 1 + 2 - 7 - 12 = -16 < 0
\]

4. **Test the function's behavior at the endpoints**:
- As \( x \to \infty \), \( f(x) \to \infty \).
- As \( x \to -\infty \), \( f(x) \to -\infty \).

5. **Use the Intermediate Value Theorem**:
- We found that:
- \( f(-\frac{7}{3}) > 0 \)
- \( f(1) < 0 \)
- There is at least one root in the interval \((- \frac{7}{3}, 1)\).

6. **Behavior at Critical Points**:
- \( f'(x) = 0 \) at \( x = -\frac{7}{3} \) and \( x = 1 \) (we confirmed \( f(-\frac{7}{3}) > 0 \) and \( f(1) < 0 \)).
- Checking the sign of \( f'(x) \):
- \( f'(x) < 0 \) for \( x < -\frac{7}{3} \),
- \( f'(x) > 0 \) for \( -\frac{7}{3} < x < 1 \) (increasing),
- \( f'(x) > 0 \) for \( x > 1 \), and \( f(x) \to \infty \).

From these observations:
1. The function changes from negative to positive between 1 and 0, indicating a root.
2. The function increases to positive after dropping below zero at \( x = 1 \).
3. Thus, the analysis shows there are three changes in sign:
- Hence, the equation \( x^3 + 2x^2 - 7x - 12 = 0 \) has **three real solutions**.

**Final Answer**: \( \boxed{3} \) real solutions.