Quadratic equations can have:

• Two different real solutions
• One repeated real solution (two equal solutions)
• No real solutions (two complex solutions)

The solutions are also called:

• Roots
• Zeros
• x-intercepts (when graphed)

Method 1: Solving by Factoring

When the quadratic can be factored, use the Zero Product Property:

If ab = 0, then a = 0 or b = 0

Steps:

1. Write in standard form
2. Factor the quadratic
3. Set each factor equal to zero
4. Solve each equation
5. Check solutions

Example 1: Solve x² + 5x + 6 = 0

Factor: (x + 2)(x + 3) = 0

Set each factor to zero: x + 2 = 0 or x + 3 = 0 x = -2 or x = -3

Solutions: x = -2 or x = -3

Check x = -2: (-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0 ✓ Check x = -3: (-3)² + 5(-3) + 6 = 9 - 15 + 6 = 0 ✓

Example 2: Solve x² - 9 = 0

Factor (difference of squares): (x + 3)(x - 3) = 0

x + 3 = 0 or x - 3 = 0 x = -3 or x = 3

Solutions: x = ±3

Example 3: Solve 2x² + 5x - 3 = 0

Factor: (2x - 1)(x + 3) = 0

2x - 1 = 0 or x + 3 = 0 x = 1/2 or x = -3

Solutions: x = 1/2 or x = -3

Example 4: Solve x² - 6x + 9 = 0

Factor (perfect square): (x - 3)² = 0

x - 3 = 0 x = 3

Solution: x = 3 (repeated root)

Method 2: Solving by Square Roots

For equations of the form x² = k, take the square root of both sides.

Remember: √(x²) = ±x (two solutions: positive and negative)

Example 1: Solve x² = 25

Take square root: x = ±√25 x = ±5

Solutions: x = 5 or x = -5

Example 2: Solve x² = 7

x = ±√7

Solutions: x = √7 or x = -√7

Example 3: Solve x² - 16 = 0

Add 16: x² = 16 x = ±4

Example 4: Solve 3x² = 75

Divide by 3: x² = 25 x = ±5

Example 5: Solve x² = -9

x = ±√(-9)

Since we can't take the square root of a negative number (in real numbers), there is no real solution.

Solving (x - h)² = k

When the equation is in the form (x - h)² = k:

Steps:

1. Take square root of both sides: x - h = ±√k
2. Solve for x: x = h ± √k

Example 1: Solve (x - 3)² = 16

Take square root: x - 3 = ±4

Two equations: x - 3 = 4 or x - 3 = -4 x = 7 or x = -1

Example 2: Solve (x + 2)² = 9

x + 2 = ±3

x + 2 = 3 or x + 2 = -3 x = 1 or x = -5

Example 3: Solve (x - 5)² = 12

x - 5 = ±√12 x - 5 = ±2√3 x = 5 ± 2√3

Solutions: x = 5 + 2√3 or x = 5 - 2√3

When to Use Each Method

Use Factoring when:

• The quadratic factors nicely with integers
• You recognize special patterns (difference of squares, perfect square trinomial)
• a, b, c are small integers

Use Square Roots when:

• The equation is x² = k or (x - h)² = k
• There's no x term (b = 0)
• The equation is already isolated

Use Quadratic Formula when:

• The quadratic doesn't factor easily
• You need exact decimal answers
• Other methods don't work

Factoring Strategy Review

For x² + bx + c = 0: Find two numbers that multiply to c and add to b

For ax² + bx + c = 0 (a ≠ 1): Use AC method or trial and error

Difference of squares: x² - k² = (x + k)(x - k)

Perfect square trinomial: x² ± 2kx + k² = (x ± k)²

Solving with GCF First

Always factor out the GCF before using other methods.

Example 1: Solve 2x² + 8x = 0

Factor out 2x: 2x(x + 4) = 0

Set each factor to zero: 2x = 0 or x + 4 = 0 x = 0 or x = -4

Example 2: Solve 3x² - 12x = 0

Factor: 3x(x - 4) = 0

3x = 0 or x - 4 = 0 x = 0 or x = 4

Example 3: Solve 5x² = 20x

Standard form: 5x² - 20x = 0 Factor: 5x(x - 4) = 0

x = 0 or x = 4

Warning: Never divide both sides by x! You'll lose the solution x = 0.

Equations Not in Standard Form

Example 1: Solve x² + 7x = -12

Standard form: x² + 7x + 12 = 0 Factor: (x + 3)(x + 4) = 0 Solutions: x = -3 or x = -4

Example 2: Solve 2x² = 5x + 3

Standard form: 2x² - 5x - 3 = 0 Factor: (2x + 1)(x - 3) = 0 Solutions: x = -1/2 or x = 3

Example 3: Solve (x + 3)(x - 1) = 5

First expand: x² + 2x - 3 = 5 Standard form: x² + 2x - 8 = 0 Factor: (x + 4)(x - 2) = 0 Solutions: x = -4 or x = 2

Important: Don't set x + 3 = 5 and x - 1 = 5! Must equal zero for zero product property.

Applications: Area Problems

Example 1: A rectangle has length 3 more than width. Area is 40. Find dimensions.

Let w = width Then w + 3 = length

Area: w(w + 3) = 40 w² + 3w = 40 w² + 3w - 40 = 0 (w + 8)(w - 5) = 0

w = -8 or w = 5

Since width must be positive: w = 5 cm Length = 8 cm

Example 2: A square has area 144 cm². Find side length.

s² = 144 s = ±12

Since side length is positive: s = 12 cm

Applications: Number Problems

Example: The product of two consecutive integers is 72. Find the integers.

Let n = first integer Then n + 1 = second integer

n(n + 1) = 72 n² + n = 72 n² + n - 72 = 0 (n + 9)(n - 8) = 0

n = -9 or n = 8

Two solutions:

• If n = -9, then integers are -9 and -8
• If n = 8, then integers are 8 and 9

Both pairs work!

Applications: Projectile Motion

Height formula: h = -16t² + v₀t + h₀

where:

• h = height at time t
• v₀ = initial velocity
• h₀ = initial height

Example: A ball is thrown upward at 48 ft/s from height 6 ft. When does it hit the ground?

h = -16t² + 48t + 6

Set h = 0 (ground level): -16t² + 48t + 6 = 0

Divide by -2: 8t² - 24t - 3 = 0

This doesn't factor nicely - would use quadratic formula.

But if it were: -16t² + 48t = 0 Factor: -16t(t - 3) = 0 t = 0 or t = 3

At t = 0 (start) and t = 3 seconds (lands)

Checking Solutions

Always substitute back into the original equation.

Example: Verify x = 2 is a solution to x² - 5x + 6 = 0

(2)² - 5(2) + 6 = 4 - 10 + 6 = 0 ✓

Common Mistakes

1. Forgetting ± when taking square roots x² = 9 has TWO solutions: x = 3 and x = -3

2. Dividing by variable Never divide both sides by x - you'll lose solutions!

3. Not setting equal to zero Must have 0 on one side to use factoring

4. Arithmetic errors Check your factoring by multiplying back

5. Forgetting negative solutions Both positive and negative roots are valid

Quick Reference

Practice Strategy

Level 1: x² = k

• x² = 16
• x² = 50

Level 2: GCF factoring

• x² + 5x = 0
• 2x² - 8x = 0

Level 3: Simple factoring

• x² + 7x + 12 = 0
• x² - 9 = 0

Level 4: Harder factoring

• 2x² + 5x + 3 = 0
• x² + 6x + 9 = 0

Level 5: Applications

• Area problems
• Number problems
• Projectile motion

Tips for Success

• Always write in standard form first
• Try factoring before other methods
• Factor out GCF when possible
• Check both solutions
• Show all work clearly
• Practice recognizing factorable patterns
• Remember the ± when taking square roots

Step 1: Calculate the discriminant $b^2 - 4ac = (6)^2 - 4(1)(2) = 36 - 8 = 28$

Step 2: Substitute into the formula $x = \frac{-6 \pm \sqrt{28}}{2}$

Step 3: Simplify $\sqrt{28}$ $\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}$

$x = \frac{-6 \pm 2\sqrt{7}}{2} = \frac{2(-3 \pm \sqrt{7})}{2} = -3 \pm \sqrt{7}$

Answer: $x = -3 + \sqrt{7}$ or $x = -3 - \sqrt{7}$

$b^2 - 4ac = (4)^2 - 4(1)(5) = 16 - 20 = -4$

Since the discriminant is negative, the equation has no real solutions (it has 2 complex solutions).

Answer: No real solutions