Choosing an Effective Strategy to Solve Quadratic Equations

8.5.3 Choosing an Effective Strategy to Solve Quadratic Equations

8.5.3 • Choosing an Effective Strategy to Solve Quadratic Equations

Activity

Solve each equation. Be prepared to explain or show your reasoning.

1.

x 2 = 121 x 2 = 121

Compare your answer:

x = 11 x = 11 and x = − 11 x = − 11

2.

x 2 − 31 = 5 x 2 − 31 = 5

Compare your answer:

x = 6 x = 6 and x = − 6 x = − 6

3.

( x − 4 ) ( x − 4 ) = 0 ( x − 4 ) ( x − 4 ) = 0

Compare your answer:

x = 4 x = 4

4.

( x + 3 ) ( x − 1 ) = 5 ( x + 3 ) ( x − 1 ) = 5

Compare your answer:

x = − 4 x = − 4 and x = 2 x = 2

5.

( x + 1 ) 2 = − 4 ( x + 1 ) 2 = − 4

Compare your answer:

No solution

6.

( x − 4 ) ( x − 1 ) = 990 ( x − 4 ) ( x − 1 ) = 990

Compare your answer:

x = 34 x = 34 and x = − 29 x = − 29

Video: Choosing an Effective Strategy to Solve Quadratic Equations

Watch the following video to learn more about choosing an effective strategy to solve quadratic equations.

Self Check

Solve the equation using any method covered in the lesson.

( x − 6 ) ( x + 5 ) = − 30.25

  1. x = 1.5
  2. x = 0.5 and x = 1.5
  3. x = 0.5
  4. No solutions

Additional Resources

Choosing an Effective Strategy to Solve Quadratic Equations

You have learned various strategies to solve quadratic equations in different forms. If possible, try to use reasoning before graphing an equation.

Let's look at some examples.

Using Reasoning and the Square Root Property

Equations such as these can be solved by reasoning and using the square root property.

Example 1

x 2 = 81 x 2 = 81

x = 9 x = 9 and x = − 9 x = − 9

Example 2

x 2 + 7 = 56 x 2 + 7 = 56

x 2 = 49 x 2 = 49

x = 7 x = 7 and x = − 7 x = − 7

Example 3

( x + 1 ) 2 = 9 ( x + 1 ) 2 = 9

x = 2 x = 2 and x = − 4 x = − 4

Each equation above is different in some way, but they all can be solved using the same process.

Using Reasoning and the Zero Product Property

Equations such as these can be solved by reasoning and using the zero product property.

Example 1

( x − 4 ) ( x + 5 ) = 0 ( x − 4 ) ( x + 5 ) = 0

x = 4 x = 4 and x = − 5 x = − 5

Example 2

( x + 3 ) ( 2 x + 7 ) = 0 ( x + 3 ) ( 2 x + 7 ) = 0

x = − 3 x = − 3 and x = − 7 2 x = − 7 2

Using Graphing to Solve

Some equations might require graphing to be solved. One method is to rearrange the equation such that it is in the form of expression = 0 = 0 . Then graph y y =expression, and find the x x -intercepts to identify the solutions.

Given: ( x + 2 ) ( x − 1 ) = 4 ( x + 2 ) ( x − 1 ) = 4

Step 1 - Rewrite the equation to set it equal to 0.

( x + 2 ) ( x − 1 ) − 4 = 0 ( x + 2 ) ( x − 1 ) − 4 = 0

Step 2 - Graph the equation with y y = the expression that equals zero.

Graph y = ( x + 2 ) ( x − 1 ) − 4 y = ( x + 2 ) ( x − 1 ) − 4 .

Graph of a parabola that opens up and passes through the points (negative 3, 0) and (2, 0).

Step 3 - Identify the x x -intercepts from the graph.

This graph has two x x -intercepts at ( − 3 , 0 ) ( − 3 , 0 ) and ( 2 , 0 ) ( 2 , 0 ) .

The original equation ( x + 2 ) ( x − 1 ) = 4 ( x + 2 ) ( x − 1 ) = 4 has two solutions at x = − 3 x = − 3 and x = 2 x = 2 .

Try it

Choosing an Effective Strategy to Solve Quadratic Equations

Solve each equation using any method.

1. ( x + 1 ) 2 + 6 = 22 ( x + 1 ) 2 + 6 = 22

Here is how to choose an effective strategy and use it to solve quadratic equations:

( x + 1 ) 2 + 6 = 22 ( x + 1 ) 2 + 6 = 22
( x + 1 ) 2 = 16 ( x + 1 ) 2 = 16
x = − 5 x = − 5 and x = 3 x = 3

2. ( x + 3 ) ( x − 8 ) = 0 ( x + 3 ) ( x − 8 ) = 0

Here is how to choose an effective strategy and use it to solve quadratic equations:

( x + 3 ) ( x − 8 ) = 0 ( x + 3 ) ( x − 8 ) = 0
x = − 3 x = − 3 and x = 8 x = 8