Lesson 1.4: Quadratic Equation Solving Methods

Master Quadratic Equation Solving Techniques

Unlock the complete toolkit for solving quadratic equations! Learn four powerful methods: square root method, completing the square, quadratic formula, and factoring. Understand when to use each method and how to analyze solutions using the discriminant.

Learning Objectives

Solve using square root method

Master completing the square

Apply quadratic formula

Analyze discriminant and solutions

Core Concept 1: Square Root Method

When to Use

Use the square root method when the quadratic equation is in the form (x + h)² = k or x² = k.

Example: $x^2=16$

Step 1:Take square root of both sides

\sqrt{x^2}=\sqrt{16}

Step 2:Simplify:

|x|=4

Step 3:Solve:

x=\pm 4

Solutions:

x=4

x=-4

Example: $(x-3)^2=25$

Step 1:Take square root:

\sqrt{(x-3)^2}=\sqrt{25}

Step 2:Simplify:

|x-3|=5

Step 3:Solve:

x-3=5

x-3=-5

Step 4:

x=8

x=-2

Perfect Square Form

(x + h)² = k

Direct square root application

Two Solutions

Always consider ± when taking square root

x = ±√k

Core Concept 2: Completing the Square

Transforming to Perfect Square

Complete the square to transform any quadratic equation into perfect square form, then apply the square root method.

Example: $x^2-6x+5=0$

Step 1:Move constant to right:

x^2-6x=-5

Step 2:Find value to complete square:

(6/2)^2=9

Step 3:Add to both sides:

x^2-6x+9=-5+9

Step 4:Factor left side:

(x-3)^2=4

Step 5:Take square root:

x-3=\pm 2

Step 6:Solve:

x=3\pm2

→

x=5

x=1

Example: 2x² + 8x - 3 = 0

Step 1:Divide by leading coefficient: x² + 4x - 3/2 = 0

Step 2:Move constant: x² + 4x = 3/2

Step 3:Complete square: (4/2)² = 4

Step 4:x² + 4x + 4 = 3/2 + 4 = 11/2

Step 5:(x + 2)² = 11/2

Step 6:x + 2 = ±√(11/2) → x = -2 ± √(11/2)

Formula for Completing Square

For x² + bx, add (b/2)²

Creates perfect square trinomial

Always Add to Both Sides

Maintain equation balance

Keep equation equivalent

Core Concept 3: Quadratic Formula & Discriminant

The Universal Method

The quadratic formula works for any quadratic equation in standard form ax² + bx + c = 0.

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Where:

• a = coefficient of x²

• b = coefficient of x

• c = constant term

Example: $2x^2+5x-3=0$

Identify:

a=2,\ b=5,\ c=-3

Substitute:

x=\dfrac{-5 \pm \sqrt{25-4\cdot2\cdot(-3)}}{4}

Simplify:

x=\dfrac{-5 \pm \sqrt{25+24}}{4}

Calculate:

x=\dfrac{-5 \pm \sqrt{49}}{4}

Result:

x=\dfrac{-5 \pm 7}{4}

Solutions:

x=1/2

x=-3

Discriminant Analysis

Δ = b² - 4ac > 0

Two distinct real solutions

Δ = b² - 4ac = 0

One repeated real solution

Δ = b² - 4ac < 0

No real solutions (complex)

Core Concept 4: Factoring Method

When Factoring Works Best

Use factoring when the quadratic can be easily factored into two binomials. This method is often the fastest when applicable.

Example: 2x² + 5x - 3 = 0

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

Step 2:Apply zero product property

Step 3:2x - 1 = 0 → x = 1/2

Or: x + 3 = 0 → x = -3

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

Example: $x^2-4=0$

Factor as difference of squares:

x^2-4=(x-2)(x+2)=0

Solutions: x = 2 or x = -2

Advantages

Fastest when applicable

No complex calculations needed

Limitations

Not always possible

May require trial and error

Real-World Application: Garden Design & Optimization

Problem Scenario

A rectangular garden has a length that is 3 meters longer than its width. The area of the garden is 28 square meters. Find the dimensions of the garden.

Solution Steps:

Step 1:Define variables: Let width = x, length = x + 3

Step 2:Write area equation: x(x + 3) = 28

Step 3:Expand: x² + 3x = 28

Step 4:Standard form: x² + 3x - 28 = 0

Step 5:Factor: (x + 7)(x - 4) = 0

Step 6:Solve: x = -7 or x = 4

Step 7:Reject negative: x = 4

Answer:Width = 4m, Length = 7m

Key Insights:

• Quadratic equations model many geometric problems
• Always check if solutions make sense in context
• Negative solutions are often rejected in real-world problems
• Multiple solution methods can verify the same answer

Practice Problems

Problem 1:

Solve using square root method: $(x+2)^2=9$

Show Solution

$\sqrt{(x+2)^2}=\sqrt{9}$

$x+2=\pm 3$

x + 2 = 3 → x = 1

x + 2 = -3 → x = -5

Solutions: x = 1 or x = -5

Problem 2:

Solve using quadratic formula: $x^2-4x+1=0$

Show Solution

$a=1,\ b=-4,\ c=1$

$x=\dfrac{4 \pm \sqrt{16-4}}{2}$

$x=\dfrac{4 \pm \sqrt{12}}{2}$

$x=\dfrac{4 \pm 2\sqrt{3}}{2}$

$x=2 \pm \sqrt{3}$

Problem 3:

A rectangle has area 15 and perimeter 16. Find its dimensions.

Show Solution

Let width = x, length = y

xy = 15, 2x + 2y = 16 → x + y = 8

y = 8 - x

x(8 - x) = 15

8x - x² = 15

x² - 8x + 15 = 0

(x - 3)(x - 5) = 0

Dimensions: 3 × 5

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Master Quadratic Equation Solving Techniques

Learning Objectives

Core Concept 1: Square Root Method

When to Use

Example: x2=16x^2=16x2=16

Example: (x−3)2=25(x-3)^2=25(x−3)2=25

Perfect Square Form

Two Solutions

Core Concept 2: Completing the Square

Transforming to Perfect Square

Example: x2−6x+5=0x^2-6x+5=0x2−6x+5=0

Example: 2x² + 8x - 3 = 0

Formula for Completing Square

Always Add to Both Sides

Core Concept 3: Quadratic Formula & Discriminant

The Universal Method

Quadratic Formula

Example: 2x2+5x−3=02x^2+5x-3=02x2+5x−3=0

Discriminant Analysis

Core Concept 4: Factoring Method

When Factoring Works Best

Example: 2x² + 5x - 3 = 0

Example: x2−4=0x^2-4=0x2−4=0

Advantages

Limitations

Real-World Application: Garden Design & Optimization

Problem Scenario

Solution Steps:

Key Insights:

Practice Problems

Example: $x^2=16$

Example: $(x-3)^2=25$

Example: $x^2-6x+5=0$

Example: $2x^2+5x-3=0$

Example: $x^2-4=0$