GED Mathematical Reasoning: Quadratic Equations I – Solve By Factoring

 

Up to this point, the equations you have solved are called “linear” equations. A linear equation is an equation with a degree of one, meaning that the highest power of the variable is one. As you know, we solve linear equations by performing opposite operations to both sides of the equal sign to get the variable by itself on one side.

The topic of this chapter is solving quadratic equations. A quadratic equation is an equation with a degree of 2, meaning that the highest power of the variable is 2.

Solving a quadratic equation still means to find the values that make the equation TRUE, but the process involves more than simply performing opposite operations to both sides of the equation. This is because quadratic equations usually contain a squared term AND a linear term, which cannot be combined since they aren’t “like terms.”

With that being said, there are several ways to solve quadratic equations. In this session, we will discuss solving quadratic equations by factoring.

  • The first step of the solving process is to get all the terms on one side of the equal sign, usually the left side, and zero on the other. This is already true of the equation in our first example so we can move on to the second step.
  • Step 2 of the process is to factor the trinomial by grouping, which is also known as the “AC” method.
  • The third step is to replace the trinomial from the original equation with its factored form, set each factor equal to zero, and solve.
  • Step 4 is to check the solutions to ensure they make the original equation true. Be sure to follow the order of operations when doing so.

 

Example 1

Solve by factoring: x^2 + x - 42 = 0

 

The equation you see in example one is an example of a quadratic equation. Notice that the highest power of x is 2.

The first step of the solving process is to get all the terms on one side of the equal sign, usually the left side, and zero on the other. This is already true of the equation in our first example so we can move on to the second step.

Step 2 of the process is to factor the trinomial by grouping, which is also known as the “AC” method. To begin this process, we’ll identify a, b, and c.

x^2 + x - 42

a = 1
b = 1
c = -42

Next we’ll multiply a times c, which is equal to -42.

a \cdot c = (1)(-42) = -42

Now we need to find two numbers whose sum is 1 and whose product is -42.

-6 + 7 = 1
-6 \cdot 7 = -42

With a little thought, those numbers are -6 and 7.

x^2 + x - 42 = x^2 - 6x + 7x - 42

Now we’ll factor by grouping. We’ll factor x out of the first group of 2 terms, leaving x - 6 And we’ll factor positive 7 out of the second group of two terms, leaving x - 6

x^2 - 6x + 7x - 42 = x(x - 6) + 7(x - 6)

Factoring out the common binomial gives us our final factored form of: (x - 6)(x + 7)

The third step is to replace the trinomial from the original equation with its factored form, set each factor equal to zero, and solve. So we will replace x^2 + x - 42 with its factored form of (x - 6)(x + 7)

x^2 + x - 42 = 0 \rightarrow (x - 6)(x + 7) = 0

Then set each factor equal to zero, resulting in the two ‘mini’ linear equations:

x - 6 = 0  and  x + 7 = 0

The reason we do this is:  If the product of two quantities is equal to zero, then one or both of the quantities must be equal to zero.

Moving on, since these “mini” equations are linear, we can solve them using opposite operations.

Focusing on the first equation, we’ll add 6 to both sides which results in a solution of x = 6

For the second equation, we’ll subtract 7 from both sides resulting in a solution of x = -7

And so the two solutions to the original equation are: x = 6 and x = -7

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It is no coincidence that the degree of the equation is TWO and we ended up with TWO solutions. Quadratic equations can have at most TWO solutions.

Step 4 is to check the solutions to ensure they make the original equation true. Be sure to follow the order of operations when doing so.

First, we’ll check x = 6

(6)^2 + (6) - 42 = 0
36 + 6 -42 = 0
0 = 0 which is true

Let’s check x = -7 as well.

(-7)^2 + (-7) - 42 = 0
49 - 7 -42 = 0
0 = 0 which is true

Therefore, we can be confident that both solutions are correct.

 

Example 2

Solve by factoring: 2x^2 + 15x = 8

 

Since the highest power of x is 2, we know it is a quadratic equation and we can proceed to solve it by factoring.

In this case, we first need to subtract 8 from both sides of the equal sign. Although we can’t combine any of the terms, since they are not “like terms,” this will make it so that all the terms are on the left side of the equal sign and zero is on the other.

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Step 2 of the process is to factor the trinomial by grouping. To begin this process, we’ll identify a, b, and c.

2x^2 + 15x - 8

a = 2
b = 15
c = -42

Next we’ll multiply a times c, which is equal to -16.

a \cdot c = (2)(-8) = -16

Now we need to find two numbers whose sum is 15 and whose product is -16.

-1 + 16 = 15
-1 \cdot 16 = -16

With a bit of thought, those numbers are -1 and 16. When we rewrite the trinomial using these two numbers, we get:

2x^2 + 15x - 8 = 2x^2 - 1x + 16x - 8

Now we’ll factor by grouping. We’ll factor x out of the first group of two terms, leaving 2 x - 1 and we’ll factor positive 8 out of the second group of two terms, leaving 2x - 1.

2x^2 - 1x + 16x - 8 = x(2x - 1) + 8(2x - 1) = (2x - 1)(x + 8)

Now on to step 3 where we’ll replace the trinomial from the original equation with its factored form, set each factor equal to zero, and solve.

 2x^2 + 15x - 8 = 0 \rightarrow (2x - 1)(x + 8) = 0

2x - 1 = 0  And  x + 8 = 0

Focusing on the first equation, we’ll add 1 to both sides and then divide both sides by 2. This gives the solution x =\frac{1}{2}.

For the second equation, we’ll subtract 8 from both sides resulting in a solution of x = -8

And so the two solutions to the original equation are: x = \frac{1}{2} and x = -8

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Finally, let’s check both solutions to ensure they make the original equation true.

First, we’ll substitute x =\frac{1}{2}.

2{\frac{1}{2}}^2 + 15\frac{1}{2} = 8

2{\frac{1}{4}} + 15\frac{1}{2} = 8

\frac{2}{4} + 15\frac{1}{2} = 8

\frac{1}{2} + \frac{15}{2} = 8

\frac{16}{2} = 8

8 = 8 which is true

Let’s check x = -8 as well.

2(-8)^2 + 15(-8) = 8

2{\frac{1}{4}} + 15\frac{1}{2} = 8

2(64) + 15(-8) = 8

128 - 120 = 8

8 = 8 which is true

 

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