SAT Math Lesson Plan 16: Solving Quadratic Equations

Lesson Summary

This 45-minute SAT math lesson focuses on solving quadratic equations using multiple algebraic strategies. Students will learn how to use the quadratic formula, complete the square, and interpret the discriminant to determine the nature of the roots. These methods are essential for solving equations that don’t factor easily—a common challenge on the SAT. The lesson includes a warm-up activity, five step-by-step teaching examples, a review section, and a 10-question multiple-choice quiz to reinforce key concepts. This lesson is part of a comprehensive 35-lesson series designed to build SAT math readiness.

Lesson Objectives

Solve quadratic equations using the quadratic formula
Complete the square to solve quadratic equations
Use the discriminant to determine the number and type of solutions
Identify the most efficient method to solve a given quadratic

Common Core Standards

HSA-REI.B.4: Solve quadratic equations in one variable
HSA-SSE.B.3b: Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines

Prerequisite Skills

Basic understanding of quadratic equations and their standard form
Ability to factor simple quadratic expressions
Comfort with simplifying radicals and working with algebraic fractions

Key Vocabulary

Quadratic Equation – An equation in the form \( ax^2 + bx + c = 0 \)
Discriminant – The expression \( b^2 - 4ac \) inside the square root of the quadratic formula
Quadratic Formula – A formula used to solve any quadratic equation: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Completing the Square – A method to transform a quadratic into a perfect square trinomial
Real Solutions – Solutions that are real numbers (as opposed to imaginary)

Warm Up

This Warm-Up helps students review solving quadratic equations using factoring—a method they should already be familiar with. This will prepare them for the alternative strategies they’ll encounter in this lesson.

Activity 1: Factoring Review

Instructions: Solve the quadratic equation by factoring and applying the zero-product property.

Problem: Solve \( x^2 + 5x + 6 = 0 \)

Step 1: Identify two numbers whose product is 6 and whose sum is 5

We need to find two numbers:

Whose product is 6 (the constant term)
Whose sum is 5 (the coefficient of the middle term)

Factor pairs of 6:

1 and 6 → 1 + 6 = 7 ❌
2 and 3 → 2 + 3 = 5 ✅

So we use 2 and 3 to factor:

\[ x^2 + 5x + 6 = (x + 2)(x + 3) \]

Step 2: Apply the zero-product property

\[ (x + 2)(x + 3) = 0 \Rightarrow x = -2 \text{ or } x = -3 \]

Answer: \( x = -2 \), \( x = -3 \)

Self-Study Tip: Always check the factor pairs of the constant. If no pair has the correct sum, the quadratic may not factor easily—and you’ll need to use the quadratic formula or complete the square.

Activity 2: Recognizing When Factoring Doesn't Work

Problem: Solve \( x^2 + x + 1 = 0 \)

Step 1: Look for two numbers whose product is 1 and whose sum is 1

Let’s test the factor pairs of 1:

1 and 1 → 1 + 1 = 2 ❌
-1 and -1 → -1 + (-1) = -2 ❌

No pair of integers has a product of 1 and a sum of 1.

Conclusion: This quadratic cannot be factored using integers. You’ll need to use the quadratic formula or complete the square to solve it.

Self-Study Tip: Always test both positive and negative factor pairs. If none satisfy both the product and sum conditions, factoring is not the best method to use.

Teach

Some quadratic equations factor easily—but many do not. That’s why the SAT often tests your ability to use more general strategies for solving quadratics, like the quadratic formula or completing the square. These techniques always work, even when factoring doesn’t.

This lesson will walk you through these strategies:

Solving using the quadratic formula
Completing the square
Using the discriminant to understand the nature of the solutions

Each example will show you how to apply the method and explain when it’s best to use it. On the SAT, problems won’t tell you which method to use—so knowing how to choose the best one is part of your success strategy.

Example 1: Solving Using the Quadratic Formula

Problem: Solve \( x^2 + 4x + 1 = 0 \)

Step 1: Identify coefficients

This quadratic is in standard form: \( ax^2 + bx + c = 0 \), where:

\( a = 1 \)
\( b = 4 \)
\( c = 1 \)

Step 2: Use the quadratic formula

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute values: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 4}}{2} \] \[ x = \frac{-4 \pm \sqrt{12}}{2} \]

Step 3: Simplify the radical

\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \] So: \[ x = \frac{-4 \pm 2\sqrt{3}}{2} \]

Step 4: Simplify the expression

\[ x = -2 \pm \sqrt{3} \]

Answer: \( x = -2 + \sqrt{3} \) or \( x = -2 - \sqrt{3} \)

Why use this method? This equation doesn’t factor easily, and completing the square would take longer. The quadratic formula is ideal for this kind of problem.

Example 2: Solving by Completing the Square

Problem: Solve \( x^2 + 6x + 5 = 0 \) by completing the square.

Step 1: Move the constant to the other side

\[ x^2 + 6x = -5 \]

Step 2: Complete the square

Take half of the coefficient of \( x \), square it, and add it to both sides. \[ \left( \frac{6}{2} \right)^2 = 9 \] Add 9 to both sides: \[ x^2 + 6x + 9 = -5 + 9 \Rightarrow (x + 3)^2 = 4 \]

Step 3: Solve by taking square roots

\[ x + 3 = \pm \sqrt{4} = \pm 2 \]

Step 4: Solve for \( x \)

\[ x = -3 + 2 = -1 \quad \text{or} \quad x = -3 - 2 = -5 \]

Answer: \( x = -1 \), \( x = -5 \)

When is completing the square useful? This method is helpful when asked to rewrite a quadratic in vertex form or when the quadratic doesn’t factor easily. It’s also essential to know because it’s how the quadratic formula is derived.

Teacher Tip: While completing the square is not commonly required on the SAT, it's the method used to rewrite a quadratic in vertex form—so it's worth knowing in case a question asks you to find a vertex algebraically.

Example 3: Using the Discriminant

Problem: Determine the number and type of solutions for \( 2x^2 + 4x + 3 = 0 \) using the discriminant.

Step 1: Identify coefficients

\[ a = 2, \quad b = 4, \quad c = 3 \]

Step 2: Use the discriminant formula

\[ D = b^2 - 4ac \] Substitute the values: \[ D = 4^2 - 4(2)(3) = 16 - 24 = -8 \]

Step 3: Interpret the result

The discriminant is negative:

If \( D > 0 \), there are two real and distinct solutions.
If \( D = 0 \), there is one real repeated solution.
If \( D < 0 \), there are no real solutions—only two complex solutions.

Conclusion: Since \( D = -8 \), the equation has two complex (non-real) solutions.

Why is this important? On the SAT, you may be asked how many real solutions a quadratic has. You don’t need to solve it—just evaluate the discriminant!

Example 4: Real-World Problem Using the Quadratic Formula

Problem: A ball is thrown upward from a platform 16 feet high. Its height in feet after \( t \) seconds is modeled by the equation:

\[ h(t) = -16t^2 + 32t + 16 \]

When will the ball hit the ground?

Step 1: Set the height to 0

\[ 0 = -16t^2 + 32t + 16 \]

Step 2: Identify coefficients

\[ a = -16, \quad b = 32, \quad c = 16 \]

Step 3: Use the quadratic formula

\[ t = \frac{-32 \pm \sqrt{32^2 - 4(-16)(16)}}{2(-16)} \] \[ t = \frac{-32 \pm \sqrt{1024 + 1024}}{-32} = \frac{-32 \pm \sqrt{2048}}{-32} \]

Step 4: Simplify the square root

\[ \sqrt{2048} = \sqrt{1024 \cdot 2} = 32\sqrt{2} \] \[ t = \frac{-32 \pm 32\sqrt{2}}{-32} \]

Step 5: Simplify

\[ t = -1 \pm \sqrt{2} \]

Since time can't be negative, we discard the negative solution:

\[ t = -1 + \sqrt{2} \approx 0.41 \text{ seconds} \]

Answer: The ball hits the ground approximately 0.41 seconds after being thrown.

Why use the quadratic formula? This equation doesn’t factor easily, and the context demands precision. The quadratic formula works every time and provides both exact and approximate solutions.

Example 5: Solving a Quadratic with One Real Solution

Problem: Solve \( x^2 - 10x + 25 = 0 \)

Step 1: Look for two numbers whose product is 25 and whose sum is -10

Let’s list factor pairs of 25:

5 and 5 → 5 + 5 = 10 ❌
-5 and -5 → -5 + (-5) = -10 ✅

Since both factors are the same, we recognize this as a perfect square trinomial:

\[ x^2 - 10x + 25 = (x - 5)^2 \]

Step 2: Set the square equal to zero

\[ (x - 5)^2 = 0 \]

Step 3: Solve for \( x \)

\[ x - 5 = 0 \Rightarrow x = 5 \]

Optional: Confirm using the quadratic formula

\[ a = 1, \quad b = -10, \quad c = 25 \Rightarrow x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(25)}}{2(1)} = \frac{10 \pm \sqrt{0}}{2} = \frac{10}{2} = 5 \]

Answer: \( x = 5 \)

Why this matters: Recognizing perfect square trinomials can save you time. When both the sum and product conditions are met with equal values, you're looking at a square of a binomial—and that means a single real solution.

Teacher Tip: Another way to confirm that a quadratic is a perfect square trinomial is to check that the discriminant is zero. For the equation \( x^2 - 10x + 25 = 0 \), the discriminant is:

\[ D = (-10)^2 - 4(1)(25) = 100 - 100 = 0 \]

This confirms the equation has exactly one real solution—a hallmark of a perfect square trinomial.

Review

This lesson showed you how to solve any quadratic equation using one of three methods: factoring (if possible), completing the square, or using the quadratic formula. You also learned how to use the discriminant to predict the number and type of solutions before solving. These strategies are critical for the SAT, especially when a quadratic doesn't factor easily.

Key takeaways:

Use factoring when the trinomial factors easily.
Use the quadratic formula when factoring is difficult or the equation contains irrational or complex roots.
Use the discriminant to determine how many solutions the equation has and whether they are real or complex.
Completing the square is useful for rewriting quadratics and for solving equations that can’t be factored.

Example 1: Solve Using the Quadratic Formula

Problem: Solve \( 3x^2 - 2x - 1 = 0 \)

\[ a = 3, \quad b = -2, \quad c = -1 \] \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \] \[ x = 1 \quad \text{or} \quad x = -\frac{1}{3} \]

Answer: \( x = 1 \), \( x = -\frac{1}{3} \)

Example 2: Use the Discriminant

Problem: How many real solutions does \( 4x^2 + 12x + 9 = 0 \) have?

Step 1: Identify coefficients

\[ a = 4, \quad b = 12, \quad c = 9 \]

Step 2: Find the discriminant

\[ D = b^2 - 4ac = 12^2 - 4(4)(9) = 144 - 144 = 0 \]

Conclusion: The discriminant is zero, which means the equation has exactly one real solution.

Additional Insight: This quadratic is also a perfect square trinomial:

\[ 4x^2 + 12x + 9 = (2x + 3)^2 \]

So the equation becomes:

\[ (2x + 3)^2 = 0 \Rightarrow x = -\frac{3}{2} \]

This confirms what the discriminant told us: one real, repeated root.

Example 3: Solve by Completing the Square

Problem: Solve \( x^2 - 4x + 1 = 0 \) by completing the square.

Step 1: Move the constant to the other side

We start by isolating the quadratic and linear terms on the left side. \[ x^2 - 4x = -1 \]

Step 2: Complete the square

Take half of the coefficient of \( x \), then square it. This value is added to both sides of the equation.

Coefficient of \( x \) is -4
Half of -4 is -2
Square of -2 is 4

Now add 4 to both sides: \[ x^2 - 4x + 4 = -1 + 4 \Rightarrow (x - 2)^2 = 3 \]

Step 3: Take the square root of both sides

\[ x - 2 = \pm \sqrt{3} \]

Step 4: Solve for \( x \)

\[ x = 2 \pm \sqrt{3} \]

Answer: \( x = 2 + \sqrt{3} \), \( x = 2 - \sqrt{3} \)

Self-Study Tip: Completing the square is especially helpful when the equation doesn't factor and you want to avoid the quadratic formula. It also prepares you to work with vertex form equations.

Multimedia Resources

For additional videos, tutorials, and interactives to reinforce the concepts in this lesson, explore this curated set of resources from Media4Math.

Click here to access the Lesson 16 Multimedia Resources.

Quiz

Directions: Solve each problem and choose the correct answer. Show all work on a separate sheet of paper if needed.

Solve for \( x \): \( x^2 - 2x - 15 = 0 \)
a) 5, -3
b) -5, 3
c) 3, 5
d) -3, -5
Solve for \( x \): \( x^2 + 4x + 8 = 0 \)
a) \( -2 \pm \sqrt{4} \)
b) \( -2 \pm \sqrt{8} \)
c) \( -2 \pm 2i \)
d) \( -4 \pm \sqrt{8} \)
Use the discriminant to determine how many real solutions the equation \( 5x^2 - 6x + 1 = 0 \) has.
a) Two
b) One
c) None
d) Cannot be determined
Solve for \( x \): \( 2x^2 + x - 1 = 0 \)
a) \( \frac{-1 \pm \sqrt{9}}{4} \)
b) \( \frac{-1 \pm \sqrt{1}}{4} \)
c) \( \frac{1 \pm \sqrt{9}}{4} \)
d) \( \frac{-1 \pm \sqrt{17}}{4} \)
Which equation has exactly one real solution?
a) \( x^2 + 2x + 1 = 0 \)
b) \( x^2 + 4x + 5 = 0 \)
c) \( x^2 + x + 1 = 0 \)
d) \( x^2 - 5x + 4 = 0 \)
What is the value of the discriminant for \( x^2 + 2x + 5 = 0 \)?
a) -16
b) 16
c) -4
d) 4
Complete the square to solve: \( x^2 + 6x + 5 = 0 \)
a) \( x = -3 \pm \sqrt{4} \)
b) \( x = -3 \pm \sqrt{9} \)
c) \( x = -3 \pm \sqrt{2} \)
d) \( x = -1, -5 \)
Which equation is easiest to solve by completing the square?
a) \( x^2 + 7x + 12 = 0 \)
b) \( x^2 - 8x + 16 = 0 \)
c) \( x^2 + 3x - 10 = 0 \)
d) \( x^2 + x - 6 = 0 \)
What type of solutions does \( x^2 + 2x + 1 = 0 \) have?
a) Two real solutions
b) One real solution
c) Two complex solutions
d) No solution
A projectile is launched with height function \( h(t) = -16t^2 + 32t \). When does it return to the ground?
a) 0 seconds
b) 1 second
c) 2 seconds
d) 4 seconds

Answer Key

Answer: a) 5, -3
Factor: \( x^2 - 2x - 15 = (x - 5)(x + 3) \)
Solutions: \( x = 5 \), \( x = -3 \)
Answer: c) \( -2 \pm 2i \)
Discriminant = -16 ⇒ \( x = -2 \pm 2i \)
Answer: a) Two
Discriminant = \( 36 - 20 = 16 \) ⇒ Two real solutions
Answer: a) \( \frac{-1 \pm \sqrt{9}}{4} \)
Quadratic formula gives exact match.
Answer: a) \( x^2 + 2x + 1 = 0 \)
Perfect square: \( (x + 1)^2 \)
Answer: a) -16
Discriminant: \( 4 - 20 = -16 \)
Answer: d) \( x = -1, -5 \)
Completing the square gives \( (x + 3)^2 = 4 \Rightarrow x = -1, -5 \)
Answer: b) \( x^2 - 8x + 16 = 0 \)
Perfect square trinomial: \( (x - 4)^2 \)
Answer: b) One real solution
Discriminant = 0 ⇒ one repeated root
Answer: c) 2 seconds
\( -16t(t - 2) = 0 \Rightarrow t = 0, 2 \) → Returns at \( t = 2 \)