Algebra 1 End-of-Course Review

Module Nine: Quadratic Equations

In this module, you will review the steps to solve quadratic equations by graphing, factoring, and using the quadratic formula.

A quadratic equation is an equation of the form y = ax² + bx + c, where “a”, “b” and “c” are real numbers, and “a” cannot be zero. It is important to note that ALL quadratic equations have a degree of 2degree of 2: In other words the equation must have an x² term, that is why a ≠ 0..

Here are some examples of quadratic equations and non-quadratic equations.

• Quadratic Equations
• Non-Quadratic Equations

Select each equation to find out why it is a quadratic equation.

This equation has an x² term.

Although it does not look like this has an x² term, look what happens when you distribute the 6x.
y = 6x² + 18x – 108
Now do you see the x² term?

This equation has an x² term.

By multiplying the two binomials, you will get:
y = (x)(x) + (x)(–1) + (1)(x) + (1)(–1)
y = x² – x + x – 1
y = x² – 1
Now you can see that this equation has
an x² term.

Select each equation to find out why it is not a quadratic equation.

This equation does not have an x² term. This equation has a degree of 1, and is called a linear equation.

Hang on here! This has an x² term, so why is it listed as a non-quadratic equation? Look what happens when you multiply the binomials and rearrange
the equation into standard form.

y + x²

=

(x)(x) + (x)(6) + (3)(x) + (3)(6)

y +

=

x² + 6x + 3x + 18

–

   –x²

y

=

9x + 18

Notice the x² terms canceled and what’s left is a linear equation.

This equation has a degree of 4.

This equation has a degree of 3.

Quadratic equations in standard form are written like: y = ax² + bx + c. For the following examples, write the quadratic equation in standard form, and then identify the values of the a, b, and c terms. Check your answers when you’re done.

Graphing a Quadratic Equation

When graphing a quadratic equation, there are two helpful pieces of information that come directly from the equation itself: the leading coefficient tells you which way the parabola will open and tells you the x-coordinate of the vertex. These provide you with a good idea of what the parabola will look like when graphed.

First, make sure the quadratic equation is in standard form y = ax² + bx + c.

• Opens Up - Minimum
• Opens Down - Maximum

If “a” is positive, the parabola opens up and has a minimum point.

If “a” is negative, the parabola opens down and has a maximum point.

Vertex –This is the maximum or minimum point on the parabola, also known as the turning point, and written in the form (x, y).

The x-coordinate of the vertex is found by substituting the values of a and b into the formula

x = .

The y-coordinate of the vertex is then found by substituting the x-coordinate of the vertex into the original quadratic equation and solving for y.

Quadratic Equations and Their Solutions

No matter which method you use, finding the solutions of the equation means that you are setting the equation to zero and finding the values of x. The solutions to the quadratic equation correspond to the x-intercepts of the parabola.

Let’s look at an example of an equation that is solved using all three methods so you can see how they relate to one another.

Example
Solve x² + 2x – 3 = 0

Solving Graphically

Solving by Factoring

Solving the equation using the Quadratic Formula

Now that you have had the opportunity to work through one whole example using each of the three methods, let’s look at some additional examples using factoring and the quadratic formula.

Solving Real-World Problems

Real-world problems, such as those involving area and projectile objects, can be solved using quadratic equations.

Area Problems

Example

Arial has an 8 x 10 photo that she would like to frame. She would like the total area of the photo and frame to equal 143 square inches. If the width of the frame is the same on all sides, what is the width of the frame? Answer

Projectile Problems

The quadratic equation for projectiles in feet and seconds is

H(t) = –16t² + vt + s, where

H(t) is the object’s current height at any time after the object is thrown,
t is the time, in seconds, the object has been in the air,
v is the object’s starting upward velocity (speed), and
s is the object’s starting height.

Example

A ball is thrown at a speed of 24 feet per second from a height of 22 feet.

1. What is the maximum height the ball will reach?
2. After how many seconds will it hit the ground?

Step 1: Write a quadratic equation in standard form to represent the situation.

Look in the example to find the values of “v” and “s” to substitute into the equation.

The initial speed is 24 feet per second, so this is the value of “v.” The ball’s starting height is 22 feet, so this is the value of “s.”

H(t) = –16t² + 24t + 22

Identify the values of a, b, and c from the quadratic equation.

H(t) = –16t² + 24t + 22, so a = –16, b = 24, and c = 22

Note: This equation does not have any x’s or y’s. In this case y is represented by H(t) which is the height in feet. x is represented by t which is time in seconds.

Step 2: Find the maximum height the ball will reach.

Recall that the maximum height is the y-value of the vertex of the graph of the equation.

Begin by finding the x-value of the vertex. x =

x =

x =

x = 0.75 This is the time (t) in seconds when it reaches the maximum height. Substitute the value of x into the “t” value of the quadratic equation and solve for H(t).

H(t) = –16t² + 24t + 22

H(t) = –16(0.75)² + 24(0.75) + 22

H(t) = –16(0.5625) + 24(0.75) + 22

H(t) = –9 + 18 + 22

H(t) = 31 feet

The ball will reach a maximum height of 31 feet.

Step 3: Find out how long it will take for the ball to hit the ground.

When the ball hits the ground, the height of the ball will be 0 feet. Substitute this value for H(t) and
solve for t.

0 = –16t² + 24t + 22

Substitute the values of a, b, and c from the equation into the quadratic formula, and solve for t.

H(t) = –16t² + 24t + 22, so a = –16, b = 24, and c = 22

t

=

t

=

t

=

t

=

 t ≈

 t ≈     and     t ≈

 t ≈ –0.642     and     t ≈ 2.142

Since time cannot be a negative number, the amount of time that it takes the ball to hit the ground is about 2 seconds.

Look here for an example that uses graphing technology to find the solutions of a quadratic equation and see how analyzing the graph of the equation can give you vital information!