8.9Solving Quadratic Equations by Square Roots
Learn when and how to apply the square root property to solve quadratic equations efficiently — including both positive and negative solutions, radical simplification, and equations with no real solutions.
Why This Matters
The square root method is the fastest way to solve quadratic equations that have no bx term. It appears on the SAT and ACT in both algebra and geometry contexts, and it is the foundation for solving vertex-form equations in Precalculus and Physics.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Quick Answer
How Do You Solve a Quadratic Equation by Square Roots?
Isolate the squared expression, take the square root of both sides, and include both the positive and negative square roots. Then solve for the variable and check the solutions in the original equation.
Basic form
x² = k
x = ±√k
Shifted form
(x − h)² = k
x = h ± √k
Learning Goals
By the end of this lesson you will be able to: (1) identify when the square root method is the most efficient approach; (2) apply the square root property to solve equations of the form ax² = k and a(x − h)² = k; (3) simplify radical solutions by factoring out perfect squares; (4) recognize equations with no real solutions; and (5) choose the best method among square roots, factoring, completing the square, and the quadratic formula.
Key Vocabulary
Square Root Property
If x² = k, then x = ±√k. Taking the square root of both sides of an equation introduces both a positive and a negative solution.
Example: x² = 25 → x = ±5
Radical
An expression that uses a root symbol (√). A square root radical √k represents the non-negative number whose square equals k.
Example: √49 = 7
Principal Square Root
The non-negative square root of a number. √k always refers to the principal (positive) root; the negative root is written −√k.
Example: √16 = 4 (not −4)
Radicand
The expression under the radical symbol. In √(3x + 1), the radicand is 3x + 1.
Example: In √20, the radicand is 20
Irrational Number
A real number that cannot be expressed as a ratio of two integers. Square roots of non-perfect-square integers are irrational.
Example: √5, √7, √11
No Real Solution
When the radicand is negative (k < 0), x² = k has no real solution because no real number squared gives a negative result.
Example: x² = −9 → no real solution
When to Use the Square Root Method
The square root method works when the equation has no bx term — that is, the equation is in one of these forms:
Use square roots
- ax² = k
- ax² + c = 0
- a(x − h)² = k
- (x − h)² + c = 0
Do NOT use square roots
- x² + 5x − 6 = 0 (has bx)
- 2x² − 3x = 0 (has bx)
- x² + 4x + 4 = 0 (has bx)
The Square Root Property
The Square Root Property — Two Solutions
x² = k
Positive root
x = +√k
Negative root
x = −√k
x = ±√k
Key: When you take the square root of both sides, you must include both the positive and negative roots. The ± symbol captures both solutions at once.
Solving a(x − h)² = k — Step by Step
Start
a(x − h)² = k
Step 1: Divide both sides by a
(x − h)² = k/a
Step 2: Take square root of both sides
x − h = ±√(k/a)
Step 3: Add h to both sides
x = h ± √(k/a)
Note: If k/a is negative, there is no real solution. If k/a = 0, there is exactly one solution: x = h.
Symmetric Solutions on the Number Line
For (x − h)² = k, the two solutions x = h ± √k are symmetric about x = h (the vertex x-coordinate).
Worked Examples
Solve x² = 49
Take the square root of both sides
x = ±√49
x = ±7
Solve x² = 20 and simplify the radical
x = ±√20
Factor the radicand: √20 = √(4 · 5)
= √4 · √5 = 2√5
Solve (x − 3)² = 16
Take the square root of both sides
x − 3 = ±√16 = ±4
x = 3 + 4 = 7 or x = 3 − 4 = −1
Solve 2(x + 1)² = 18
Divide both sides by 2
(x + 1)² = 9
x + 1 = ±3
x = −1 + 3 = 2 or x = −1 − 3 = −4
Solve 3x² − 12 = 0
Add 12 to both sides: 3x² = 12
Divide by 3: x² = 4
x = ±√4 = ±2
Solve (x + 2)² = −9
Take the square root of both sides
x + 2 = ±√(−9)
The square root of a negative number is not real
Solve (1/2)(x − 5)² = 8
Multiply both sides by 2: (x − 5)² = 16
x − 5 = ±4
x = 5 + 4 = 9 or x = 5 − 4 = 1
Solve x² = 7. Leave in exact form and verify one solution.
x = ±√7
Verify x = √7: (√7)² = 7 ✓
Verify x = −√7: (−√7)² = 7 ✓
Example 9 — Method Comparison
Why should x² + 5x − 6 = 0 NOT be solved by the square root method?
This equation has a bx term (5x). The square root method requires isolating a perfect square, but x² + 5x cannot be written as (x + h)² without completing the square first. Instead, factor directly:
x² + 5x − 6 = 0
(x + 6)(x − 1) = 0
x = −6 or x = 1
Rule: If the equation has a bx term, use factoring, completing the square, or the quadratic formula — not the square root method.
Choosing a Quadratic-Solving Method
Use this table to decide which method is most efficient for a given equation.
| Method | Use When | Example |
|---|---|---|
| Square Roots | No bx term: ax² = k or a(x−h)² = k | 3x² = 27 or (x−2)² = 9 |
| Factoring | Factorable trinomial; integer roots expected | x² − 5x + 6 = 0 → (x−2)(x−3) |
| Completing the Square | Convert to vertex form; a = 1 preferred | x² + 6x + 2 = 0 |
| Quadratic Formula | Always works; use when other methods are inconvenient | 2x² − 3x − 7 = 0 |
Common Mistakes
x² = 25 → x = 5 only (forgot negative root)
x² = 25 → x = ±5; both +5 and −5 satisfy the equation because (−5)² = 25.
Taking the square root before isolating: √(2x²) = √18 → x = 3
Isolate first: 2x² = 18 → x² = 9 → x = ±3. Always divide by the coefficient before taking the square root.
√(a + b) = √a + √b (e.g., √(9 + 16) = 3 + 4 = 7)
√(9 + 16) = √25 = 5. You cannot split a square root over addition. Simplify inside the radical first.
x² = 20 → x = √20 ≈ 4.47 (not simplified)
x = ±√20 = ±2√5. Simplify by factoring out perfect squares: √20 = √(4 · 5) = 2√5.
x² = −4 → x = ±2i (treating as real)
x² = −4 has no real solution. Stop here unless complex numbers are required by the problem.
Losing a solution: (x − 3)² = 16 → x − 3 = 4 → x = 7 only
x − 3 = ±4 gives two solutions: x = 7 and x = −1. Always use ± when taking the square root.
Applying square roots to x² + 5x = 9 directly
This has a bx term — use factoring, completing the square, or the quadratic formula instead.
Forgetting to check solutions in the original equation
Always substitute back to verify, especially when simplifying radicals or working with shifted forms.
Guided Practice
Guided Practice Video: Solving Quadratic Equations by Square Roots
Review solving quadratic equations by isolating the squared term and taking square roots — including ± solutions — before completing the guided problems below.
Video by Sang Real Math
Watch on YouTube ↗Solve x² = 64.
Hint: Take the square root of both sides. Remember to include both the positive and negative roots.
Solve (x + 5)² = 36.
Hint: Take the square root of both sides: x + 5 = ±6. Then solve for x in each case.
Solve 4(x − 2)² = 100.
Hint: Divide both sides by 4 first to get (x − 2)² = 25. Then take the square root.
Solve x² = 12 and simplify the radical.
Hint: x = ±√12. Factor 12 = 4 · 3. Then √12 = √4 · √3 = 2√3.
Practice Questions
Interactive Practice — 5 Questions
Solve x² = 36.
Solve (x − 5)² = 9.
Which equation is best solved by the square root method?
Solve x² = −25.
Simplify √20.