Precalculus Guidance: Radical, Absolute Value, and Inequality Equations

Q1. Solve: $x + \sqrt{x - 12} = 0$

Background

Topic: Radical Equations

This question tests your ability to solve equations involving square roots (radicals). You'll need to isolate the radical and use algebraic techniques to solve for $x$.

Key Terms and Formulas:

• Radical equation: An equation in which the variable is inside a root.

• To solve, isolate the radical and then square both sides to eliminate the root.

Step-by-Step Guidance

1. Isolate the radical term $\sqrt{x - 12}$ on one side of the equation.

2. Express the equation so that $\sqrt{x - 12}$ is alone.

3. Square both sides to eliminate the square root.

4. Solve the resulting quadratic equation for $x$.

Try solving on your own before revealing the answer!

Final Answer: $x = 4$

After isolating the radical and squaring both sides, you get $x - 12 = x^2$. Solving this quadratic gives $x = 4$ as the valid solution (after checking for extraneous roots).

Q2. Solve: $3\sqrt{x - 3} + 2 = 0$

Background

Topic: Radical Equations

This question involves solving an equation with a square root. You'll need to isolate the radical and then square both sides.

Key Terms and Formulas:

• Radical equation: Variable inside a root.

• Isolate the radical, then square both sides.

Step-by-Step Guidance

1. Subtract 2 from both sides to isolate the radical term.

2. Divide both sides by 3 to further isolate $\sqrt{x - 3}$.

3. Square both sides to eliminate the square root.

4. Solve for $x$ and check for extraneous solutions.

Try solving on your own before revealing the answer!

Final Answer: $x = 3$

After isolating and squaring, you find $x - 3 = 0$, so $x = 3$ is the solution. Always check for extraneous solutions in radical equations.

Q3. Solve: $2(\sqrt{x - 3})^2 - 5\sqrt{x - 3} - 3 = 0$

Background

Topic: Radical Equations (Quadratic in form)

This equation is quadratic in terms of $\sqrt{x - 3}$. You'll use substitution to solve.

Key Terms and Formulas:

• Let $u = \sqrt{x - 3}$, then rewrite the equation in terms of $u$.

• Quadratic formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Step-by-Step Guidance

1. Let $u = \sqrt{x - 3}$ and rewrite the equation as $2u^2 - 5u - 3 = 0$.

2. Use the quadratic formula to solve for $u$.

3. Once you have $u$, substitute back $u = \sqrt{x - 3}$.

4. Solve for $x$ by squaring both sides if needed.

Try solving on your own before revealing the answer!

Final Answer: $x = 4$ and $x = 8$

Solving the quadratic gives $u = 6$ and $u = -0.5$. Only $u = 6$ is valid for a square root, so $x = 39$ and $x = 3.25$ (after checking for extraneous solutions).

Q4. Solve and graph the inequality: $-5 \leq 3x - 2 < 13$

Background

Topic: Compound Inequalities

This question tests your ability to solve and graph compound inequalities, and express the solution in interval notation.

Key Terms and Formulas:

• Compound inequality: Two inequalities joined together.

• Interval notation: Expresses the solution set as an interval.

Step-by-Step Guidance

1. Add 2 to all parts of the inequality to isolate the $3x$ term.

2. Divide all parts by 3 to solve for $x$.

3. Express the solution in interval notation.

4. Graph the interval on a number line, using closed or open circles as appropriate.

Try solving on your own before revealing the answer!

Final Answer: $[-1, 5)$

The solution is $x$ between $-1$ and $5$, including$-1$but not $5$. Graph this as a closed circle at $-1$ and open at $5$.

Q5. Solve and graph the inequality: $-4x - 17 < 33$

Background

Topic: Linear Inequalities

This question tests your ability to solve a linear inequality and express the answer in interval notation.

Key Terms and Formulas:

• Linear inequality: An inequality involving a linear expression.

• Interval notation: Expresses the solution set as an interval.

Step-by-Step Guidance

1. Add 17 to both sides to isolate the $-4x$ term.

2. Divide both sides by $-4$ (remember to reverse the inequality sign when dividing by a negative).

3. Express the solution in interval notation.

4. Graph the interval on a number line.

Try solving on your own before revealing the answer!

Final Answer: $(-12.5, \infty)$

After solving, you get $x > -12.5$. Graph this as an open circle at $-12.5$ and shade to the right.

Q6. Solve and graph the inequality: $3x + 7 \leq -1$ or $x + 7 > 12$

Background

Topic: Compound (OR) Inequalities

This question tests your ability to solve two inequalities joined by "or" and express the solution in interval notation.

Key Terms and Formulas:

• "Or" means the solution is any value that satisfies either inequality.

• Interval notation: Use union ($\cup$) to combine intervals.

Step-by-Step Guidance

1. Solve $3x + 7 \leq -1$ for $x$.

2. Solve $x + 7 > 12$ for $x$.

3. Express each solution in interval notation.

4. Combine the intervals using union ($\cup$).

5. Graph the solution set on a number line.

Try solving on your own before revealing the answer!

Final Answer: $(-\infty, -\frac{8}{3}] \cup (5, \infty)$

The solution is all $x \leq -\frac{8}{3}$ or $x > 5$. Graph as two intervals.

Q7. Solve the equation: $|3x - 8| = 17$

Background

Topic: Absolute Value Equations

This question tests your ability to solve equations involving absolute value.

Key Terms and Formulas:

• Absolute value equation: $|A| = B$ means $A = B$ or $A = -B$.

Step-by-Step Guidance

1. Set up two equations: $3x - 8 = 17$ and $3x - 8 = -17$.

2. Solve each equation for $x$.

3. Check both solutions in the original equation.

Try solving on your own before revealing the answer!

Final Answer: $x = 25/3$ and $x = -3$

Solving both equations gives two solutions. Both satisfy the original absolute value equation.

Q8. Solve and graph the inequality: $|2x + 1| > 9$

Background

Topic: Absolute Value Inequalities

This question tests your ability to solve inequalities involving absolute value and express the answer in interval notation.

Key Terms and Formulas:

• For $|A| > B$, the solution is $A < -B$ or $A > B$.

• Interval notation: Use union ($\cup$) to combine intervals.

Step-by-Step Guidance

1. Set up two inequalities: $2x + 1 > 9$ and $2x + 1 < -9$.

2. Solve each inequality for $x$.

3. Express each solution in interval notation.

4. Combine the intervals using union ($\cup$).

5. Graph the solution set on a number line.

Try solving on your own before revealing the answer!

Final Answer: $(-\infty, -5) \cup (4, \infty)$

The solution is $x < -5$ or $x > 4$. Graph as two intervals.

Q9. Solve and graph the inequality: $|5x - 4| + 2 \leq 18$

Background

Topic: Absolute Value Inequalities

This question tests your ability to solve inequalities involving absolute value and express the answer in interval notation.

Key Terms and Formulas:

• For $|A| \leq B$, the solution is $-B \leq A \leq B$.

Step-by-Step Guidance

1. Subtract 2 from both sides to isolate the absolute value term.

2. Set up the compound inequality: $-16 \leq 5x - 4 \leq 16$.

3. Solve for $x$ in the compound inequality.

4. Express the solution in interval notation.

5. Graph the interval on a number line.

Try solving on your own before revealing the answer!

Final Answer: $(-2.4, 4)$

The solution is $x$ between $-2.4$ and $4$, including both endpoints. Graph as a closed interval.

Q10. Solve and graph the inequality: $2|x + 6| - 3 < -7$

Background

Topic: Absolute Value Inequalities

This question tests your ability to solve inequalities involving absolute value and express the answer in interval notation.

Key Terms and Formulas:

• Isolate the absolute value term before solving.

• For $|A| < B$, the solution is $-B < A < B$.

Step-by-Step Guidance

1. Add 3 to both sides to isolate $2|x + 6|$.

2. Divide both sides by 2 to isolate $|x + 6|$.

3. Set up the inequality $|x + 6| < -2$.

4. Consider whether this inequality has any real solutions (absolute value is always non-negative).

Try solving on your own before revealing the answer!

Final Answer: No solution

Since absolute value cannot be less than a negative number, there are no real solutions to this inequality.