Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter3: Multi-step Equations And Inequalities
Section: Chapter Questions
Problem 11PTTS
Related questions
Question
![# Algebra Problem Set
Welcome to the Algebra Problem Set section. Below are some example problems and their respective solutions. Follow along to understand the methods used to solve these algebraic equations.
### Example Problems and Solutions:
**Problem 1:**
Solve for \( x \),
\[
x = \left( \sqrt[3]{8} \right)^2
\]
**Solution:**
First, find the cube root of 8,
\[
\sqrt[3]{8} = 2
\]
Next, square the result,
\[
2^2 = 4
\]
Therefore,
\[
x = 4
\]
---
**Problem 2:**
Solve for \( x \),
\[
(5 - x)^{1/2} - 2x = 0
\]
**Solution:**
This problem combines a square root and a linear term. Rearrange and solve step by step:
\[
(5 - x)^{1/2} = 2x
\]
Square both sides to eliminate the square root,
\[
5 - x = (2x)^2
\]
\[
5 - x = 4x^2
\]
Rearrange the equation,
\[
4x^2 + x - 5 = 0
\]
Solve for \( x \) using the quadratic formula where \( a = 4 \), \( b = 1 \), and \( c = -5 \),
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
\[
x = \frac{-1 \pm \sqrt{1 + 80}}{8}
\]
\[
x = \frac{-1 \pm \sqrt{81}}{8}
\]
\[
x = \frac{-1 \pm 9}{8}
\]
Thus, the solutions are:
\[
x = 1 \quad \text{or} \quad x = -\frac{5}{4}
\]
---
**Problem 3:**
Solve for \( x \),
\[
8x^{3/2} = 1000
\]
**Solution:**
First, isolate \( x \),
\[
x^{3/2} = \frac{1000}{8}
\]
\[
x^{3/2} = 125
\]
Next, take both](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff33bf94-c662-48a9-a87f-2a0124279f60%2Fcade914e-b76f-47f6-9fd0-1ab8e460d7b7%2Fsm7sezn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# Algebra Problem Set
Welcome to the Algebra Problem Set section. Below are some example problems and their respective solutions. Follow along to understand the methods used to solve these algebraic equations.
### Example Problems and Solutions:
**Problem 1:**
Solve for \( x \),
\[
x = \left( \sqrt[3]{8} \right)^2
\]
**Solution:**
First, find the cube root of 8,
\[
\sqrt[3]{8} = 2
\]
Next, square the result,
\[
2^2 = 4
\]
Therefore,
\[
x = 4
\]
---
**Problem 2:**
Solve for \( x \),
\[
(5 - x)^{1/2} - 2x = 0
\]
**Solution:**
This problem combines a square root and a linear term. Rearrange and solve step by step:
\[
(5 - x)^{1/2} = 2x
\]
Square both sides to eliminate the square root,
\[
5 - x = (2x)^2
\]
\[
5 - x = 4x^2
\]
Rearrange the equation,
\[
4x^2 + x - 5 = 0
\]
Solve for \( x \) using the quadratic formula where \( a = 4 \), \( b = 1 \), and \( c = -5 \),
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
\[
x = \frac{-1 \pm \sqrt{1 + 80}}{8}
\]
\[
x = \frac{-1 \pm \sqrt{81}}{8}
\]
\[
x = \frac{-1 \pm 9}{8}
\]
Thus, the solutions are:
\[
x = 1 \quad \text{or} \quad x = -\frac{5}{4}
\]
---
**Problem 3:**
Solve for \( x \),
\[
8x^{3/2} = 1000
\]
**Solution:**
First, isolate \( x \),
\[
x^{3/2} = \frac{1000}{8}
\]
\[
x^{3/2} = 125
\]
Next, take both
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