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High School Algebra II: Homework Help Resource26 chapters | 281 lessons | 2 flashcard sets

Instructor:
*Laura Pennington*

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

This lesson will go over the steps to solve the equation x^2 - 6x = 16. We will examine two different solving processes, and learn how to check our work for accuracy.

We will look at two methods of solving *x* 2 - 6*x* = 16. The first uses factoring, and the second uses the quadratic formula.

When solving by factoring, we will use the following steps:

1.) Get all non-zero terms on one side of the equation, and zero on the other.

2.) Factor the non-zero side of the equation.

3.) Set each factor equal to zero and solve.

The first thing we want to do is get all non-zero terms on one side of the equation. To do this, we subtract 16 from both sides of the equation as shown.

The next step is to factor the non-zero side of the equation, so we want to factor *x* 2 - 6*x* - 16. When the coefficient in front of *x* 2 is one, as it is in our case, we can use the process shown in the image to factor.

We want to factor *x* 2 - 6*x* - 16, so we want to fill the blanks of (*x* + _____)(*x* + _____) with two numbers that when multiplied, are equal to -16 and when added, are equal to -6. We can find this by listing the numbers that multiply together to equal -16, and then check each one to see if they add up to -6.

Factors of 16 | Sum |
---|---|

1 and -16 | 1 + (-16) = -15 |

-1 and 16 | -1 + 16 = 15 |

2 and -8 | 2 + (-8) = -6 |

-2 and 8 | -2 + 8 = 6 |

4 and -4 | 4 + (-4) = 0 |

The only pair of factors of 16 that add up to -6 are 2 and -8, so these are the numbers we will use to fill in the blanks. That is, *x* 2 - 6*x* - 16 = (*x* + 2)(*x* - 8).

The last step is to set each of these factors equal to zero and solve as shown.

We see that *x* = -2 or *x* = 8.

Another way to solve this problem is using the **quadratic formula**. The quadratic formula gives solutions to an equation of the form *a**x* 2 + *b**x* + *c* = 0.

To solve using the quadratic formula, we use the following steps:

1.) Put the equation in the form *a**x* 2 + *b**x* + *c* = 0 by getting all non-zero terms on one side and zero on the other side.

2.) Identify the values of *a*, *b*, and *c* in your equation from step 1.

3.) Plug your values for *a*, *b*, and *c* into the quadratic formula and simplify.

In our example, we want to solve *x* 2 - 6*x* = 16. The first step is the same as the first step when we used factoring. We get all the non-zero terms on one side by subtracting 16 from both sides to get *x* 2 - 6*x* - 16 = 0.

The next step is to identify the values of *a*, *b*, and *c* in our equation. We can recognize this by noting that *a* is the number in front of *x* 2, *b* is the number in front of *x*, and *c* is the number that is all by itself. In our example, *a* = 1, *b* = -6, and *c* = -16.

Lastly, we plug these values into our quadratic formula and simplify as shown in the following work.

Again, we see that *x* = -2 or *x* = 8.

Both of our solving processes revealed that the solution to *x* 2 - 6*x* = 16 is *x* = -2 or *x* = 8, so these are both solutions to the equation.

When solving equations of any kind, checking your work is quite easy! Simply plug the solution, or solutions, into the original equation. If it makes a true statement, then you know your solution is correct. If not, then you know you made an error somewhere along the way.

In our case, we plug each of our values for *x* that we found into *x* 2 - 6*x* = 16. If they make a true statement, then we've done everything correctly.

Wonderful! We see that our solutions check out, so we did everything correctly. The solutions to *x* 2 - 6*x* = 16 are *x* = -2 and *x* = 8.

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High School Algebra II: Homework Help Resource26 chapters | 281 lessons | 2 flashcard sets

- How to Evaluate a Polynomial in Function Notation 8:22
- Understanding Basic Polynomial Graphs 9:15
- Basic Transformations of Polynomial Graphs 7:37
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- Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11
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