College Algebra Quiz: Absolute Value Equations

What this quiz covers

This quiz focuses on Absolute Value Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $-5$ (correct answer)
2. $5$
3. $-1$
4. $1$

Explanation: From $|x - 2|^2 = 9$, we get $|x - 2| = 3$ (since absolute value is non-negative). This gives us two equations: $x - 2 = 3$ or $x - 2 = -3$. Solving: $x = 5$ or $x = -1$. The product of all solutions is $5 \times (-1) = -5$.

Question 2

1. $\{-5, -1, 1, 5\}$ (correct answer)
2. $\{-5, 1, 5\}$
3. $\{-1, 1\}$
4. $\{-5, 5\}$

Explanation: For $||x| - 3| = 2$, we first solve $|x| - 3 = 2$ or $|x| - 3 = -2$. From $|x| - 3 = 2$: $|x| = 5$, so $x = 5$ or $x = -5$. From $|x| - 3 = -2$: $|x| = 1$, so $x = 1$ or $x = -1$. Let's verify: For $x = 5$: $||5| - 3| = |5 - 3| = |2| = 2$ ✓. For $x = -5$: $||-5| - 3| = |5 - 3| = |2| = 2$ ✓. For $x = 1$: $||1| - 3| = |1 - 3| = |-2| = 2$ ✓. For $x = -1$: $||-1| - 3| = |1 - 3| = |-2| = 2$ ✓. All four solutions are valid.

Question 3

1. $x = -6$ only
2. $x = \frac{4}{5}$ only
3. $x = -6$ and $x = \frac{4}{5}$
4. $x = -\frac{4}{5}$ and $x = 6$ (correct answer)

Explanation: For $|3x - 1| = |2x + 5|$, we have two cases: Case 1: $3x - 1 = 2x + 5$, which gives $x = 6$. Case 2: $3x - 1 = -(2x + 5) = -2x - 5$, so $3x - 1 = -2x - 5$, giving $5x = -4$, so $x = -\frac{4}{5}$. Let's verify: For $x = 6$: $|3(6) - 1| = |17| = 17$ and $|2(6) + 5| = |17| = 17$ ✓. For $x = -\frac{4}{5}$: $|3(-\frac{4}{5}) - 1| = |-\frac{12}{5} - 1| = |-\frac{17}{5}| = \frac{17}{5}$ and $|2(-\frac{4}{5}) + 5| = |-\frac{8}{5} + 5| = |\frac{17}{5}| = \frac{17}{5}$ ✓. Both solutions are valid.

Question 4

1. 0
2. 1 (correct answer)
3. 2
4. infinitely many

Explanation: For $|x + 3| = 2x - 5$ to have solutions, we need $2x - 5 \geq 0$, so $x \geq \frac{5}{2}$. Case 1: If $x + 3 \geq 0$ (i.e., $x \geq -3$), then $x + 3 = 2x - 5$, giving $x = 8$. Since $8 \geq \frac{5}{2}$ and $8 \geq -3$, this solution is valid. Case 2: If $x + 3 < 0$ (i.e., $x < -3$), then $-(x + 3) = 2x - 5$, so $-x - 3 = 2x - 5$, giving $3x = 2$, so $x = \frac{2}{3}$. But $\frac{2}{3} \not< -3$, so this contradicts our case assumption. Therefore, there is exactly 1 solution: $x = 8$.