Question
(1) \( x^{2}-3 x-2=0 \)
Ask by Higgins Mann. in South Africa
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The solutions to the equation \( x^{2} - 3x - 2 = 0 \) are:
\[
x = \frac{3 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{17}}{2}
\]
Approximately:
\[
x \approx 3.5616 \quad \text{and} \quad x \approx -0.5616
\]
Solution
To solve the quadratic equation:
\[
x^{2} - 3x - 2 = 0
\]
we can use the **quadratic formula**, which is applicable to any quadratic equation of the form \( ax^{2} + bx + c = 0 \). The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
\]
### Step-by-Step Solution
1. **Identify the coefficients** in the equation \( x^{2} - 3x - 2 = 0 \):
- \( a = 1 \) (coefficient of \( x^{2} \))
- \( b = -3 \) (coefficient of \( x \))
- \( c = -2 \) (constant term)
2. **Plug the coefficients** into the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}
\]
3. **Simplify inside the square root**:
\[
(-3)^{2} = 9 \\
4 \cdot 1 \cdot (-2) = -8 \\
\]
So,
\[
x = \frac{3 \pm \sqrt{9 - (-8)}}{2} = \frac{3 \pm \sqrt{17}}{2}
\]
4. **Express the final solutions**:
\[
x = \frac{3 + \sqrt{17}}{2} \quad \text{or} \quad x = \frac{3 - \sqrt{17}}{2}
\]
### Approximate Numerical Values
If you need approximate decimal values:
\[
\sqrt{17} \approx 4.1231
\]
So,
\[
x \approx \frac{3 + 4.1231}{2} = \frac{7.1231}{2} \approx 3.5616
\]
\[
x \approx \frac{3 - 4.1231}{2} = \frac{-1.1231}{2} \approx -0.5616
\]
### Summary of Solutions
- **Exact Solutions**:
\[
x = \frac{3 + \sqrt{17}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{17}}{2}
\]
- **Approximate Decimal Solutions**:
\[
x \approx 3.5616 \quad \text{and} \quad x \approx -0.5616
\]
These are the two solutions to the quadratic equation \( x^{2} - 3x - 2 = 0 \).
Reviewed and approved by the UpStudy tutoring team
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Mind Expander
To solve the quadratic equation \( x^{2}-3x-2=0 \), you can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a=1 \), \( b=-3 \), and \( c=-2 \). Plugging in these values gives us \( b^2-4ac = 9 + 8 = 17 \). Thus, the solutions are \( x = \frac{3 \pm \sqrt{17}}{2} \). This equation finds its roots at approximately \( x \approx 4.561 \) and \( x \approx -1.561 \). Quadratics like this model countless real-world situations, from projectile motion to financial growth scenarios, highlighting just how essential they are in mathematics and everyday life!