Chapter 8: Problem 59
Rationalize each denominator. Write quotients in lowest terms. $$ \frac{\sqrt{7}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} $$
Short Answer
Expert verified
The rationalized form is \( \root{21} + \root{14} + \root{6} + 2 \).
Step by step solution
01
- Identify the Conjugate
To rationalize the denominator, identify the conjugate of the denominator. The conjugate of \(\root{3} - \root{2}\) is \(\root{3} + \root{2}\).
02
- Multiply Numerator and Denominator by the Conjugate
Multiply both the numerator and the denominator by the conjugate \(\root{3} + \root{2}\): \[ \frac{\root{7} + \root{2}}{\root{3} - \root{2}} \times \frac{\root{3} + \root{2}}{\root{3} + \root{2}}. \] This results in \[ \frac{(\root{7} + \root{2})(\root{3} + \root{2})}{(\root{3} - \root{2})(\root{3} + \root{2})}. \]
03
- Expand the Numerator
Expand the numerator using the distributive property: \[ \root{7}\root{3} + \root{7}\root{2} + \root{2}\root{3} + \root{2}\root{2} = \root{21} + \root{14} + \root{6} + 2. \]
04
- Simplify the Denominator
Simplify the denominator using the difference of squares formula: \[ \root{3}^2 - \root{2}^2 = 3 - 2 = 1. \]
05
- Combine Results
Since the denominator is now 1, the expression simplifies to just the numerator: \[ \root{21} + \root{14} + \root{6} + 2. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference of Squares
The difference of squares is a special algebraic pattern used to simplify expressions. It states that for any two numbers, say \(a\) and \(b\), the expression \(a^2 - b^2\) can be factored into \((a + b)(a - b)\).
In the given exercise, the denominator \(\root{3} - \root{2}\) is not yet in a factored form. To rationalize it, we multiply by the conjugate \(\root{3} + \root{2}\).
When we do this multiplication, we utilize the difference of squares formula. Here’s how it works step-by-step:
In the given exercise, the denominator \(\root{3} - \root{2}\) is not yet in a factored form. To rationalize it, we multiply by the conjugate \(\root{3} + \root{2}\).
When we do this multiplication, we utilize the difference of squares formula. Here’s how it works step-by-step:
- Original denominator: \(\root{3} - \root{2}\)
- Conjugate: \(\root{3} + \root{2}\)
- Multiplying these: \((\root{3} - \root{2})(\root{3} + \root{2})\)
- Using the formula gives: \((\root{3})^2 - (\root{2})^2\)
- Which simplifies to: \(3 - 2 = 1\)
Distributive Property
The distributive property is a fundamental principle in algebra. It allows us to multiply a single term across terms inside parentheses. The property is written as \(a(b + c) = ab + ac\).
In our problem:
The numerator after multiplying by the conjugate is \((\root{7} + \root{2})(\root{3} + \root{2})\). Using the distributive property, we need to distribute each term:
In our problem:
The numerator after multiplying by the conjugate is \((\root{7} + \root{2})(\root{3} + \root{2})\). Using the distributive property, we need to distribute each term:
- \(\root{7} \cdot \root{3}\)
- \(\root{7} \cdot \root{2}\)
- \(\root{2} \cdot \root{3}\)
- \(\root{2} \cdot \root{2}\)
- \(\root{21}\)
- \(\root{14}\)
- \(\root{6}\)
- 2 (since \(\root{2} \cdot \root{2} = 2\))
Simplifying Radicals
Simplifying radicals is the process of making a radical expression as simple as possible. This often involves combining like terms or removing a radical from the denominator, as seen in rationalization. Let's look at what was done to simplify the terms:
In our problem, simplifying involves:
Finally, since dividing by 1 doesn’t change the value, the expression simplifies directly to \(\root{21} + \root{14} + \root{6} + 2\).
Always look to combine like terms and ensure all radicals are in their simplest forms.
In our problem, simplifying involves:
- Combining the radicals in the numerator to simplify the expression.
- Making the denominator rational, which is 1 in this case.
Finally, since dividing by 1 doesn’t change the value, the expression simplifies directly to \(\root{21} + \root{14} + \root{6} + 2\).
Always look to combine like terms and ensure all radicals are in their simplest forms.