Problem 277
Question
In the following exercises, simplify. $$ (5-\sqrt{7})(4-\sqrt{7}) $$
Step-by-Step Solution
Verified Answer
27 - 9 \( \sqrt{7} \)
1Step 1: Identify the binomials
Recognize that the expression \( (5-\backslash sqrt{7})(4-\backslash sqrt{7}) \) involves two binomials. Binomials are algebraic expressions with two terms.
2Step 2: Apply the distributive property
Apply the distributive property (FOIL method) which states \( (a-b)(c-d) = ac - ad - bc + bd \). Here, \( a = 5 \), \( b = \backslash sqrt{7} \), \( c = 4 \), and \( d = \backslash sqrt{7} \). Expand the expression: \[ (5-\backslash sqrt{7})(4-\backslash sqrt{7}) = 5 \backslash times 4 - 5 \backslash times \backslash sqrt{7} - \backslash sqrt{7} \backslash times 4 + \backslash sqrt{7} \backslash times \backslash sqrt{7} \] and simplify each term.
3Step 3: Multiply each term
Calculate each product: \[ 5 \backslash times 4 = 20 \] \[ 5 \backslash times \backslash sqrt{7} = 5 \backslash sqrt{7} \] \[ \backslash sqrt{7} \backslash times 4 = 4 \backslash sqrt{7} \] \[ \backslash sqrt{7} \backslash times \backslash sqrt{7} = 7 \]
4Step 4: Combine like terms
Combine the results from Step 3: \[ 20 - 5 \backslash sqrt{7} - 4 \backslash sqrt{7} + 7 \] Combine \( -5 \backslash sqrt{7} \) and \( -4 \backslash sqrt{7} \) to get \[ 20 + 7 - (5 \backslash sqrt{7} + 4 \backslash sqrt{7}) = 27 - 9 \backslash sqrt{7} \]
5Step 5: Write the final simplified expression
The simplified form of the expression is: \[ 27 - 9 \backslash sqrt{7} \]
Key Concepts
Distributive PropertyBinomialsFOIL MethodRadicals
Distributive Property
The distributive property is a foundational concept in algebra. This property tells us how to multiply a single term and two or more terms inside a set of parentheses.
In its simplest form, it states that for any numbers or expressions, \(a(b + c) = ab + ac\).
This means you distribute the multiplication of \(a\) to both \(b\) and \(c\), then add the results.
When dealing with binomials like \((5-\backslash\text{sqrt}{7})(4-\backslash\text{sqrt}{7})\), the distributive property helps break down and simplify the multiplication process.
You'll apply the multiplication step-by-step to make it easier. In this case, we distribute each term in the first binomial by each term in the second binomial.
In its simplest form, it states that for any numbers or expressions, \(a(b + c) = ab + ac\).
This means you distribute the multiplication of \(a\) to both \(b\) and \(c\), then add the results.
When dealing with binomials like \((5-\backslash\text{sqrt}{7})(4-\backslash\text{sqrt}{7})\), the distributive property helps break down and simplify the multiplication process.
You'll apply the multiplication step-by-step to make it easier. In this case, we distribute each term in the first binomial by each term in the second binomial.
Binomials
Binomials are algebraic expressions that consist of exactly two terms. For example, in our exercise, \(5-\backslash\text{sqrt}{7}\) and \(4-\backslash\text{sqrt}{7}\) are binomials.
Binomials can involve numbers, variables, or even radicals like in this case.
Recognizing that you are working with binomials helps you know which algebraic rules to apply, such as the FOIL method.
Another important point about binomials is that they often lead to terms that can be combined further, as seen in the final simplifying steps.
Binomials can involve numbers, variables, or even radicals like in this case.
Recognizing that you are working with binomials helps you know which algebraic rules to apply, such as the FOIL method.
Another important point about binomials is that they often lead to terms that can be combined further, as seen in the final simplifying steps.
FOIL Method
The FOIL method stands for First, Outside, Inside, Last. It's a technique used to multiply two binomials. By applying this method, you ensure that each term in the first binomial is multiplied by each term in the second binomial.
Let's apply the FOIL method to our example:
Let's apply the FOIL method to our example:
- **First**: Multiply the first terms: \(5 \times 4 = 20\).
- **Outside**: Multiply the outer terms: \(5 \times -\backslash\text{sqrt}{7} = -5\backslash\text{sqrt}{7}\).
- **Inside**: Multiply the inner terms: \(-\backslash\text{sqrt}{7} \times 4 = -4\backslash\text{sqrt}{7}\).
- **Last**: Multiply the last terms: \(-\backslash\text{sqrt}{7} \times -\backslash\text{sqrt}{7} = 7\).
Radicals
A radical is simply an expression that involves the root of a number, such as the square root \(\backslash\text{sqrt}{7}\). Radicals often emerge within algebraic problems and need specialized handling.
For instance, \(\backslash\text{sqrt}{7} \times \backslash\text{sqrt}{7} = 7\) because the square root of any number times itself returns the original number.
In our example, the radical term \(-\backslash\text{sqrt}{7}\) shows up multiple times during the multiplication process.
Understanding how to simplify expressions involving radicals is key to achieving the correct final answer in algebra problems.
By seeing how radicals combine and simplify, you can confidently manage them within more complex expressions.
For instance, \(\backslash\text{sqrt}{7} \times \backslash\text{sqrt}{7} = 7\) because the square root of any number times itself returns the original number.
In our example, the radical term \(-\backslash\text{sqrt}{7}\) shows up multiple times during the multiplication process.
Understanding how to simplify expressions involving radicals is key to achieving the correct final answer in algebra problems.
By seeing how radicals combine and simplify, you can confidently manage them within more complex expressions.
Other exercises in this chapter
Problem 275
In the following exercises, simplify. $$ (9-\sqrt{2})(6+\sqrt{2}) $$
View solution Problem 276
In the following exercises, simplify. $$ (3-\sqrt{7})(5-\sqrt{7}) $$
View solution Problem 278
In the following exercises, simplify. $$ (1+3 \sqrt{10})(5-2 \sqrt{10}) $$
View solution Problem 279
In the following exercises, simplify. $$ (7-2 \sqrt{5})(4+9 \sqrt{5}) $$
View solution