Topics
8. Quadratic Equations
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United States of AmericaVirginia
Algebra 1

8.01 Solve quadratics using graphs and tables

Lesson
Practice

Solving quadratic equations using graphs and tables

A quadratic equation is a polynomial equation of degree 2. The solutions to a quadratic equation are the values that make the equation true. The solutions are also the ordered pairs that make up all of the points on the curve.

A table with 3 columns titled x, x squared equals f of x, and Point, and with 7 rows. The data is as follows: First row: negative 2.8, quantity negative 2.8 squared equals 7.84, (negative 2.8, 7.84); Second row: negative 2, quantity negative 2 squared equals 4, (negative 2, 4); Third row: negative 3 halves, quantity negative 3 halves squared equals 9 over 4, (negative 3 halves, 9 over 4); Fourth row: 0, 0 squared equals 0, (0, 0); Fifth row: 2 thirds, 2 thirds squared equals 4 over 9, (2 thirds, 9 over 4); Sixth row: 2.4, 2.4 squared equals 5.76, (2.4, 5.76); Seventh row: 3, 3 squared equals 9, (3, 9).
Some solutions to f(x)=x^2 are shown in the table
The graph of f of x equals x squared plotted on a first and second quadrant coordinate plane. The graph is shown passing through the following points: (-2.8, 7.84), (-2, 4), (-1.5, 2.25), (0, 0), (0.67, 0.44), (2.4, 5.76), and (3, 9).
Solutions to f(x)=x^2 are any point on the curve

The solutions to a quadratic equation where y is equal to zero are the x-intercepts of the corresponding function. They are also known as the roots of the equation or the zeros of the function.

A table with 2 columns titled x and f(x), and with 7 rows. The data is as follows: First row: negative 3, 9; Second row: negative 2, 4; Third row: negative 1, 1; Fourth row: 0, 0; Fifth row: 1, 1; Sixth row: 2, 4; Seventh row: 3, 9. The fourth row containing 0, 0 is labeled Solution to x squared equals 0.
Table of f(x)=x^2
he graph of f(x) equals x squared plotted on a first and second quadrant coordinate plane. The point (0, 0) is labeled Solution to x squared equals 0.
Graph of f(x)=x^2

In the graph and table shown, we see x=0 is the only solution. This is because x=0 is the only value that makes x^2=0 true.

If we tried to find the solution to x^2=-2 there would be no real solutions, because squaring any non-zero real number will give a positive result.

The roots, or zeros, in a quadratic function occur when f(x)=0.

The method we will use to solve a problem such as x^2=4 is by creating an equivalent equation by rearranging it so it is equal to 0 and then identifying the x-intercepts.

\displaystyle x^2\displaystyle =\displaystyle 4Given equation
\displaystyle x^2-4\displaystyle =\displaystyle 0Subtract 4 from both sides

Next we can replace the 0 in the equation with y to get y=x^2-4. The graph of this equation is the graph of y=x^2 shifted down 4 units so the graph of y=x^2-4 is:

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
f(x)

We can see that the graph crosses the x-axis at -2 and 2, so the solutions to x^2=4 are -2 and 2.

We can check this using substitution: \left(-2\right)^2=4 \text{ and } \left(2\right)^2=4

We can follow this process to solve any quadratic equation graphically. In other words, for any function f(x)=c, for some real number constant, c, we can write the equivalent equation f(x)-c=0, and find the x-intercepts of g(x) = f(x) - c to solve for x.

A quadratic equation can have one, two or no real solutions.

x
y
One real solution
x
y
Two real solutions
x
y
No real solutions

Examples

Example 1

Solve the equation 2x^2 = 18.

Worked Solution
Create a strategy

We can write an equivalent equation set equal to zero, and then use a table to find the zeros of the new function.

Apply the idea

If we set this equation equal to zero, we would get

\displaystyle 2x^2\displaystyle =\displaystyle 18Given equation
\displaystyle 2x^2-18\displaystyle =\displaystyle 0Subtraction property of equality

When building a table, we want to choose values within a suitable range so we don't have to do too many calculations. Start by finding the values in the domain -4\leq x\leq4. If the y-value (also called the function value) is zero for any of these x-values, then we have found a solution to the corresponding equation.

In the table, we are looking for the entries where y=0.

x-4-3-2-101234
y140-10-16-18-16-10014

We can see the equation has solutions of x=-3,x=3, which we can also write as x= \pm 3.

Reflect and check

We can see that the table of y-values has both positive and negative values. Whenever this is the case for a function of the form f\left(x\right)=ax^2+bx+c, we know that the equation 0=ax^2+bx+c must have two real solutions, and the corresponding parabola will have two x-intercepts.

Example 2

Consider the function y=\left(x-2\right)^2-9.

a

Draw a graph of the function.

Worked Solution
Create a strategy

The function is given in vertex form so we know the vertex is at \left(2, -9\right). We can substitute x=0 to find the y-intercept at \left(0, -5\right). We can find other points on the curve by substituting in other values, and by filling a table of values.

Apply the idea
x01234
y-5-8-9-8-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
x
-10
-8
-6
-4
-2
2
4
6
y
Reflect and check

It is important when drawing graphs to clearly show the key features such as the vertex and the intercepts by choosing appropriate scales for the axes.

b

Determine the solution(s) to the equation 0=\left(x-2\right)^2-9.

Worked Solution
Create a strategy

We can find the solution(s) by looking at the graph from part (a) and identifying where it crosses the x-axis.

Apply the idea

The solutions to the equation can be found at the x-intercepts of the graph we drew in part (a).

-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
x
-10
-8
-6
-4
-2
2
4
6
y

The solutions are x=-1 and x=5.

Reflect and check

We can verify the solutions by substituting each one into the equation for x. If the right side of the equation evaluates to 0, then it is a solution.

First, let's check the solution x=-1

\displaystyle 0\displaystyle =\displaystyle (x-2)^2 -9Original equation
\displaystyle 0\displaystyle =\displaystyle (-1-2)^2-9Substitute in x=-1
\displaystyle 0\displaystyle =\displaystyle (-3)^2-9Simplify inside parenthesis
\displaystyle 0\displaystyle =\displaystyle 9-9Evaluate the square
\displaystyle 0\displaystyle =\displaystyle 0Subtract

0=0 is a true statement, so x=-1 is a solution to 0=\left(x-2\right)^2 -9.

Now, let's check our other solution, x=5

\displaystyle 0\displaystyle =\displaystyle (x-2)^2 -9Original equation
\displaystyle 0\displaystyle =\displaystyle (5-2)^2-9Substitute in x=5
\displaystyle 0\displaystyle =\displaystyle (3)^2-9Simplify inside parenthesis
\displaystyle 0\displaystyle =\displaystyle 9-9Evaluate the square
\displaystyle 0\displaystyle =\displaystyle 0Subtract

0=0 is a true statement, so x=5 is also a solution to 0=\left(x-2\right)^2 -9.

Example 3

The graph below shows the path of a rock after it has been thrown from a cliff where x represents the time in seconds and f(x) represents the height of the rock in feet.

Height of the rock
2
4
6
8
10
x\text{ time in seconds}
4
8
12
16
20
24
28
32
36
40
f(x)\text{ height in ft}
a

Find and interpret f(8).

Worked Solution
Create a strategy

We are looking for the value of f(x), the height of the rock, when the value of x is 8 seconds.

Apply the idea

Use the graph to identify the value of f(x) when x=8

Height of the rock
2
4
6
8
10
\text{Time (}x\text{, seconds)}
4
8
12
16
20
24
28
32
36
40
\text{Height (}y\text{, ft)}

So, 8 seconds after being thrown the rock has reached a height of 26 feet.

b

Determine when the rock has a height of 38 feet.

Worked Solution
Create a strategy

We are looking for when the rock reaches a height of 38 feet. In other words, we are given the height, the value of f(x), and we are looking for the time, the x-values.

Apply the idea

Notice that there are 2 places where the graph has a y-value of 38. The solutions will be the x-values of those 2 points.

Height of the rock
2
4
6
8
10
\text{Time (}x\text{, seconds)}
4
8
12
16
20
24
28
32
36
40
\text{Height (}y\text{, ft)}

The rock reaches a height of 38 feet after 2 seconds and 6 seconds.

Reflect and check

We can use the vertex to determine the equation of this graph. The vertex is at (4, 42). The parabola is facing downard, so we know a will be negative. We can use the vertex form to write an equation:

y=a(x-4)^2+42

We can plug in the coordinates for another known point, like the y-intercept, which is (0, 26) and then solve for a. In this case, the result will be a = -1. We just found the solutions to the corresponding equation -(x-4)^2+42=38.

c

Estimate the viable solution. Explain why only one solution is viable.

Worked Solution
Create a strategy

A viable solution is a solution that makes sense in the context of the problem. Keep in mind that time cannot be negative.

Apply the idea

There is a positive x-intercept at approximately \left(10.5,0\right). This means the rock hits the ground after about \left(10.5,0\right) seconds.

Looking at the graph we can see the other x-intercept is to the left of the origin which would represent a negative value for time which is not possible. This is why there is only one viable solution.

Example 4

Identify the number of real solutions each quadratic function has.

a
-4
-3
-2
-1
1
2
3
4
x
-1
1
2
3
4
5
6
y
Worked Solution
Create a strategy

Real solutions correspond with x-intercepts. How many x-intercepts does this function have?

Apply the idea

This quadratic never crosses the x-axis so there are no x-intercepts.

The function has 0 real solutions.

b
-1
1
2
3
4
5
6
7
x
-3
-2
-1
1
2
3
4
y
Worked Solution
Create a strategy

Real solutions correspond with x-intercepts. How many x-intercepts does this function have?

Apply the idea

The quadratic crosses the x-axis in two different spots.

-1
1
2
3
4
5
6
7
x
-3
-2
-1
1
2
3
4
y

This quadratic has two real solutions.

Reflect and check

We can determine from the graph the exact value of the two real solutions. The solutions are x=2 and x=5.

Idea summary

We solve a quadratic equation by creating an equivalent equation by making your equation set equal to 0.

The solutions to a quadratic equation are any values that make the equation true.

If the equation is equal to 0, the solutions are called roots of the equation or zeros of the function. These correspond to the x-intercepts of the graph.

For any function f(x)=c, for some real number constant, c, we can write the equivalent equation f(x)-c=0, and find the x-intercepts of g(x) = f(x) - c to solve for x.

Outcomes

A.EI.3

The student will represent, solve, and interpret the solution to a quadratic equation in one variable.

A.EI.3a

Solve a quadratic equation in one variable over the set of real numbers with rational or irrational solutions, including those that can be used to solve contextual problems.

A.EI.3b

Determine and justify if a quadratic equation in one variable has no real solutions, one real solution, or two real solutions.

A.EI.3c

Verify possible solution(s) to a quadratic equation in one variable algebraically, graphically, and with technology to justify the reasonableness of answer(s). Explain the solution method and interpret solutions for problems given in context.

A.F.2

The student will investigate, analyze, and compare characteristics of functions, including quadratic and exponential functions, and model quadratic and exponential relationships.

A.F.2c

Graph a quadratic function, f(x), in two variables using a variety of strategies, including transformations f(x) + k and kf(x), where k is limited to rational values.

A.F.2g

For any value, x, in the domain of f, determine f(x) of a quadratic or exponential function. Determine x given any value f(x) in the range of f of a quadratic function. Explain the meaning of x and f(x) in context.

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