🎯 Learning Objectives
- How to solve equations with square roots in them.
- How to solve equations with variables in denominators.
- What an EXTRANEOUS solution is and why you must always check.
- How to use cross-multiplication safely.
- Strategies for SAT-style radical and rational problems.
- Practice questions with worked solutions.
1. Radical Equations
A radical equation contains a square root (or other root) of a variable expression.
The Radical Strategy
To remove a square root: ISOLATE it, then SQUARE both sides
2. The 4 Steps for Radical Equations
STEP 1: Isolate the radical on one side
STEP 2: Square both sides to eliminate the root
STEP 3: Solve the resulting equation
STEP 4: CHECK each answer in the original equation
💡 Why You MUST Check: Extraneous Solutions
Squaring both sides is a 'one-way' operation that can introduce false answers. For example, x = −3 is not the same as x² = 9, because x² = 9 also accepts x = +3.
Always plug your answer back into the ORIGINAL (un-squared) equation. If both sides match, you have a real solution. If they don't, the answer is extraneous: discard it.
📝 Example: Solve √(x + 3) = x − 3
Square both sides: x + 3 = (x − 3)² = x² − 6x + 9
Rearrange: 0 = x² − 7x + 6 = (x − 1)(x − 6)
Solutions: x = 1 or x = 6
Check x = 1: √4 = 2 vs. 1 − 3 = −2 ✗ EXTRANEOUS
Check x = 6: √9 = 3 vs. 6 − 3 = 3 ✓
Final answer: x = 6 only.
3. Rational Equations
A rational equation has variables in the denominators. Two reliable strategies:
- Cross-multiply: When you have ONE fraction equal to ONE fraction. Multiply diagonally.
- Common denominator: When there are 3 or more terms. Multiply every term by the least common denominator.
4. Cross-Multiplication
Only works for two equal fractions
a/b = c/d ⟹ a·d = b·c
📝 Example: Solve 3/(x − 1) = 6/x
Cross-multiply: 3x = 6(x − 1)
Distribute: 3x = 6x − 6
Solve: −3x = −6 → x = 2
Check: 3/(2−1) = 3, 6/2 = 3 ✓
5. Domain Restrictions
In any rational equation, denominators cannot equal zero. Note any x-values that make a denominator zero: these are off-limits, even if they emerge as 'solutions'.
6. Strategies
⚡ STRATEGY 1: Always square AFTER isolating
Isolate a radical before squaring when possible. If you square a sum, expand the entire square, including the cross term.
⚡ STRATEGY 2: Plug answer choices in for radicals
Multiple choice + radicals = a perfect chance to back-solve. Just plug each choice into the original equation.
⚡ STRATEGY 3: Cancel before cross-multiplying
If the same factor appears on both sides of a rational equation, cancel it first to simplify.
⚠️ Common Mistakes
- Forgetting to check for extraneous solutions after squaring.
- Squaring when the radical is NOT yet isolated.
- (x − 3)² becoming x² + 9 (forgetting the middle term −6x).
- Allowing a 'solution' that makes a denominator zero.
7. Summary
📌 Key Takeaways
- Radical strategy: Isolate → Square → Solve → CHECK.
- Extraneous: Squaring can create fake answers: always verify.
- Cross-multiply: a/b = c/d ⟹ ad = bc (use only with two fractions).
- More than 2 fractions: Multiply every term by the LCD.
- Forbidden: Any x that makes a denominator 0.
8. Practice: 10 Questions
Q1. Solve √x = 5.
A) 5 B) 25 C) ±25 D) 10
Hint: Square both sides: x = 25.
Q2. Solve √(x + 4) = 3.
A) 5 B) 7 C) 9 D) 13
Hint: Square: x + 4 = 9 → x = 5.
Q3. Solve 6/x = 2.
A) 2 B) 3 C) 4 D) 12
Hint: Cross-multiply: 6 = 2x → x = 3.
Q4. Solve 4/(x − 1) = 2.
A) 1 B) 2 C) 3 D) 5
Hint: Cross-multiply: 4 = 2(x − 1) → x = 3.
Q5. Solve √(2x + 1) = x − 1.
A) 0 B) 4 C) 0 and 4 D) Only 4 (0 is extraneous)
Hint: Square: 2x + 1 = x² − 2x + 1 → x² − 4x = 0. Check both.
Q6. What is the domain restriction for 5/(x − 3) = 2?
A) x ≠ 0 B) x ≠ 2 C) x ≠ 3 D) x ≠ 5
Hint: Denominator can't be 0: x ≠ 3.
Q7. Solve 3/x + 2 = 5.
A) 1 B) 2 C) 3 D) 5
Hint: Subtract 2: 3/x = 3 → x = 1.
Q8. Solve √(x − 2) + 3 = 7.
A) 6 B) 16 C) 18 D) 20
Hint: Isolate: √(x−2) = 4. Square: x − 2 = 16 → x = 18.
Q9. Cross-multiply: x/4 = 6/8.
A) 1 B) 2 C) 3 D) 4
Hint: 8x = 24 → x = 3.
Q10. Solve √x + 1 = x − 5.
A) 4 B) 9 C) 4 and 9 D) 9 (4 extraneous)
Hint: Isolate: √x = x − 6. Square: x = x² − 12x + 36 → x² − 13x + 36 = 0 → (x−4)(x−9). Check: x=4 fails, x=9 works.
Answer Key & Worked Solutions
# | Answer | Reasoning |
Q1. | B) 25 | Square both sides. |
Q2. | A) 5 | x + 4 = 9. |
Q3. | B) 3 | Cross-multiply. |
Q4. | C) 3 | Cross-multiply and solve. |
Q5. | D) Only 4 | x = 0 makes radical = 1 but RHS = −1 → extraneous. |
Q6. | C) x ≠ 3 | Denominator can't be zero. |
Q7. | A) 1 | Subtract 2 first. |
Q8. | C) 18 | Isolate then square. |
Q9. | C) 3 | Standard cross-multiply. |
Q10. | D) 9 | x = 4 fails the check. |