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The solutions to the equations are:
1. \( x = \frac{-3 \pm \sqrt{17}}{4} \) (approximately \( x \approx 0.2808 \) and \( x \approx -1.7808 \))
2. \( x = \frac{15 \pm 5\sqrt{7}}{2} \) (approximately \( x \approx 14.1146 \) and \( x \approx 0.8854 \))
3. \( x = 15 \) and \( x = 8 \)
4. \( x = \frac{3}{4} \)
Solución
Sure! Let's solve each of the quadratic equations you've provided. We'll use the quadratic formula for each, which is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
---
### 1. \( 2x^{2} + 3x - 1 = 0 \)
**Coefficients:**
- \( a = 2 \)
- \( b = 3 \)
- \( c = -1 \)
**Discriminant (\( D \)):**
\[
D = b^2 - 4ac = 3^2 - 4(2)(-1) = 9 + 8 = 17
\]
**Solutions:**
\[
x = \frac{-3 \pm \sqrt{17}}{2 \times 2} = \frac{-3 \pm \sqrt{17}}{4}
\]
**Approximate Values:**
\[
x \approx \frac{-3 + 4.1231}{4} \approx \frac{1.1231}{4} \approx 0.2808
\]
\[
x \approx \frac{-3 - 4.1231}{4} \approx \frac{-7.1231}{4} \approx -1.7808
\]
---
### 2. \( 2x^{2} - 30x + 25 = 0 \)
**Coefficients:**
- \( a = 2 \)
- \( b = -30 \)
- \( c = 25 \)
**Discriminant (\( D \)):**
\[
D = (-30)^2 - 4(2)(25) = 900 - 200 = 700
\]
**Solutions:**
\[
x = \frac{30 \pm \sqrt{700}}{2 \times 2} = \frac{30 \pm 10\sqrt{7}}{4} = \frac{15 \pm 5\sqrt{7}}{2}
\]
**Approximate Values:**
\[
\sqrt{7} \approx 2.6458
\]
\[
x \approx \frac{30 + 26.4583}{4} = \frac{56.4583}{4} \approx 14.1146
\]
\[
x \approx \frac{30 - 26.4583}{4} = \frac{3.5417}{4} \approx 0.8854
\]
---
### 3. \( x^{2} - 23x + 120 = 0 \)
**Coefficients:**
- \( a = 1 \)
- \( b = -23 \)
- \( c = 120 \)
**Discriminant (\( D \)):**
\[
D = (-23)^2 - 4(1)(120) = 529 - 480 = 49
\]
**Solutions:**
\[
x = \frac{23 \pm \sqrt{49}}{2 \times 1} = \frac{23 \pm 7}{2}
\]
**Values:**
\[
x = \frac{23 + 7}{2} = \frac{30}{2} = 15
\]
\[
x = \frac{23 - 7}{2} = \frac{16}{2} = 8
\]
---
### 4. \( 16x^{2} - 24x + 9 = 0 \)
**Coefficients:**
- \( a = 16 \)
- \( b = -24 \)
- \( c = 9 \)
**Discriminant (\( D \)):**
\[
D = (-24)^2 - 4(16)(9) = 576 - 576 = 0
\]
Since the discriminant is zero, there is **one real repeated root**.
**Solution:**
\[
x = \frac{24}{2 \times 16} = \frac{24}{32} = \frac{3}{4}
\]
**Alternatively, Factoring:**
\[
16x^2 - 24x + 9 = (4x - 3)^2 = 0 \implies 4x - 3 = 0 \implies x = \frac{3}{4}
\]
---
### Summary of Solutions
1. **\( 2x^{2} + 3x - 1 = 0 \)**
- \( x = \frac{-3 \pm \sqrt{17}}{4} \)
- Approximately \( x \approx 0.2808 \) and \( x \approx -1.7808 \)
2. **\( 2x^{2} - 30x + 25 = 0 \)**
- \( x = \frac{15 \pm 5\sqrt{7}}{2} \)
- Approximately \( x \approx 14.1146 \) and \( x \approx 0.8854 \)
3. **\( x^{2} - 23x + 120 = 0 \)**
- \( x = 15 \) and \( x = 8 \)
4. **\( 16x^{2} - 24x + 9 = 0 \)**
- \( x = \frac{3}{4} \)
If you have any further questions or need additional explanations for any of these solutions, feel free to ask!
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