Chapter 6: Problem 106
Factor the expression in part a and solve the equation in part \(\boldsymbol{b}\) a. \(x^{2}-36\) b. \(x^{2}-36=0\)
Short Answer
Expert verified
a. \((x - 6)(x + 6)\); b. \(x = 6\) and \(x = -6\).
Step by step solution
01
Understand the Type of Expression
The given expression \(x^2 - 36\) is a classic example of a difference of squares, which has the formula \(a^2 - b^2 = (a - b)(a + b)\). In this case, \(a = x\) and \(b = 6\) because \(36 = 6^2\).
02
Apply the Difference of Squares Formula
Using the difference of squares formula, we can factor \(x^2 - 36\) as follows:\[x^2 - 36 = (x - 6)(x + 6)\]
03
Set the Equation to Zero
For part b, we start with the equation \(x^2 - 36 = 0\). Using the factorization from Step 2, the equation becomes \((x - 6)(x + 6) = 0\).
04
Apply the Zero Product Property
The Zero Product Property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). Here, \((x - 6)\) and \((x + 6)\) are the factors, so set each to 0:1. \(x - 6 = 0\)2. \(x + 6 = 0\)
05
Solve for x
Solve the equations from Step 4:1. For \(x - 6 = 0\), add 6 to both sides to get \(x = 6\).2. For \(x + 6 = 0\), subtract 6 from both sides to get \(x = -6\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Squares
In algebra, the **difference of squares** is a powerful tool for factoring expressions. This concept applies when you have an expression of the form \(a^2 - b^2\). Here, \(a\) and \(b\) represent any expressions or numbers. The beauty of this technique is that it simplifies into a multiplication of two binomials: \((a - b)(a + b)\).
For example, in the expression \(x^2 - 36\), \(x^2\) can be seen as \((x)^2\) and 36 as \(6^2\). Recognizing this makes it clear that you are dealing with a difference of squares. Simply apply the formula:
For example, in the expression \(x^2 - 36\), \(x^2\) can be seen as \((x)^2\) and 36 as \(6^2\). Recognizing this makes it clear that you are dealing with a difference of squares. Simply apply the formula:
- Identify \(a\) and \(b\) from the expression such that \(a^2 = x^2\) and \(b^2 = 36\).
- Here, \(a = x\) and \(b = 6\).
- Apply the formula: \(x^2 - 36 = (x - 6)(x + 6)\).
Zero Product Property
The **Zero Product Property** is a fundamental principle in algebra that comes in handy when solving equations. It tells us that if a product of two factors equals zero, then at least one of the factors must be zero. In algebraic terms, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
When you use factoring to solve quadratic equations, such as in the equation \(x^2 - 36 = 0\), it often leads us to apply this property. Once we factor \(x^2 - 36\) into \((x - 6)(x + 6) = 0\), the next step is to set each factor equal to zero:
When you use factoring to solve quadratic equations, such as in the equation \(x^2 - 36 = 0\), it often leads us to apply this property. Once we factor \(x^2 - 36\) into \((x - 6)(x + 6) = 0\), the next step is to set each factor equal to zero:
- \(x - 6 = 0\)
- \(x + 6 = 0\)
Solving Quadratic Equations
**Solving quadratic equations** often starts with recognizing the type of quadratic and deciding on the best strategy for solving it. One effective approach is factoring, especially when dealing with equations that simplify into a form easily recognizable, such as a difference of squares.
Let's consider the quadratic equation from our exercise: \(x^2 - 36 = 0\). Here's how you solve it step by step:
Let's consider the quadratic equation from our exercise: \(x^2 - 36 = 0\). Here's how you solve it step by step:
- Recognize the form: Notice that \(x^2 - 36\) is a difference of squares which allows factoring to \((x - 6)(x + 6)\).
- Set each factor equal to zero using the Zero Product Property:
- \(x - 6 = 0\)
- \(x + 6 = 0\)
- Solve each simple equation:
- For \(x - 6 = 0\): Add 6 to both sides to find \(x = 6\).
- For \(x + 6 = 0\): Subtract 6 from both sides to find \(x = -6\).