Question
10. \( x^{\frac{1}{2}}-3 x^{\frac{1}{4}}+2= \) 43. \( x-x^{\frac{1}{2}}-20=0 \) 16. \( x-3 x^{\frac{1}{2}}-18= \) 19. \( x^{\frac{2}{3}}-3 x^{\frac{1}{3}}-4=0 \)
Ask by Bob Peters. in South Africa
Feb 02,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Solutions to the equations are:
1. \( x = 1 \) and \( x = 16 \)
2. \( x = 25 \)
3. \( x = 36 \)
4. \( x = 64 \) and \( x = -1 \)
Solution
Sure, let's solve each of the given equations step by step.
---
### **10.** \( \sqrt{x} - 3 \sqrt[4]{x} + 2 = 0 \)
**Solution:**
1. **Substitute Variable:**
Let \( y = \sqrt[4]{x} \). Then, \( \sqrt{x} = y^2 \).
2. **Rewrite the Equation:**
\[
y^2 - 3y + 2 = 0
\]
3. **Factor the Quadratic:**
\[
(y - 1)(y - 2) = 0
\]
So, \( y = 1 \) or \( y = 2 \).
4. **Find \( x \):**
\[
x = y^4
\]
- If \( y = 1 \), then \( x = 1 \).
- If \( y = 2 \), then \( x = 16 \).
**Solutions:** \( x = 1 \) and \( x = 16 \)
---
### **43.** \( x - \sqrt{x} - 20 = 0 \)
**Solution:**
1. **Substitute Variable:**
Let \( y = \sqrt{x} \). Then, \( x = y^2 \).
2. **Rewrite the Equation:**
\[
y^2 - y - 20 = 0
\]
3. **Solve the Quadratic:**
\[
y = \frac{1 \pm \sqrt{1 + 80}}{2} = \frac{1 \pm 9}{2}
\]
- \( y = 5 \) or \( y = -4 \)
4. **Discard Negative Root:**
Since \( y = \sqrt{x} \geq 0 \), \( y = 5 \).
5. **Find \( x \):**
\[
x = 5^2 = 25
\]
**Solution:** \( x = 25 \)
---
### **16.** \( x - 3\sqrt{x} - 18 = 0 \)
**Solution:**
1. **Substitute Variable:**
Let \( y = \sqrt{x} \). Then, \( x = y^2 \).
2. **Rewrite the Equation:**
\[
y^2 - 3y - 18 = 0
\]
3. **Solve the Quadratic:**
\[
y = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm 9}{2}
\]
- \( y = 6 \) or \( y = -3 \)
4. **Discard Negative Root:**
Since \( y = \sqrt{x} \geq 0 \), \( y = 6 \).
5. **Find \( x \):**
\[
x = 6^2 = 36
\]
**Solution:** \( x = 36 \)
---
### **19.** \( x^{\frac{2}{3}} - 3x^{\frac{1}{3}} - 4 = 0 \)
**Solution:**
1. **Substitute Variable:**
Let \( y = x^{\frac{1}{3}} \). Then, \( x^{\frac{2}{3}} = y^2 \).
2. **Rewrite the Equation:**
\[
y^2 - 3y - 4 = 0
\]
3. **Solve the Quadratic:**
\[
y = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}
\]
- \( y = 4 \) or \( y = -1 \)
4. **Find \( x \):**
\[
x = y^3
\]
- If \( y = 4 \), then \( x = 64 \).
- If \( y = -1 \), then \( x = -1 \).
**Solutions:** \( x = 64 \) and \( x = -1 \)
---
### **Summary of Solutions:**
1. **Problem 10:** \( x = 1 \) and \( x = 16 \)
2. **Problem 43:** \( x = 25 \)
3. **Problem 16:** \( x = 36 \)
4. **Problem 19:** \( x = 64 \) and \( x = -1 \)
---
If you have any further questions or need additional explanations, feel free to ask!
Reviewed and approved by the UpStudy tutoring team
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Explain 
Simplify this solution The Deep Dive
For the expression \( x^{\frac{1}{2}} - 3 x^{\frac{1}{4}} + 2 \), you can simplify it by performing a substitution. Letting \( y = x^{\frac{1}{4}} \), the expression transforms to \( y^2 - 3y + 2 \), which can be factored into \( (y - 1)(y - 2) \). This leads to the solutions \( y = 1 \) or \( y = 2 \), and plugging back gives \( x = 1^4 = 1 \) or \( x = 2^4 = 16 \). Now, for \( x - x^{\frac{1}{2}} - 20 = 0 \), rearrange it as \( x - 20 = x^{\frac{1}{2}} \). Squaring both sides yields a quadratic equation \( x^2 - 40x + 400 = 0 \). Factoring gives \( (x - 20)^2 = 0 \), leading to \( x = 20 \) as a double solution, indicating it touches the x-axis at this point.
