Question
Express the following surd fractions with rational denominators:
1. Root 7 minus 2, all on root 7 plus 2.
2. 3 root 6 minus 2, all on 3 root 6 plus 2.
1. Root 7 minus 2, all on root 7 plus 2.
2. 3 root 6 minus 2, all on 3 root 6 plus 2.
Answers
Answer
How did you calculate the denominator for the first question.
Answer
Thanks that was very helpful. π«ΆπΌ
Answered by
GPT-4o mini
To express the given surd fractions with rational denominators, we'll multiply both the numerator and denominator by the conjugate of the denominator.
### 1. \(\frac{\sqrt{7} - 2}{\sqrt{7} + 2}\)
The conjugate of the denominator \(\sqrt{7} + 2\) is \(\sqrt{7} - 2\).
Now we can multiply the numerator and denominator:
\[
\frac{\sqrt{7} - 2}{\sqrt{7} + 2} \cdot \frac{\sqrt{7} - 2}{\sqrt{7} - 2} = \frac{(\sqrt{7} - 2)^2}{(\sqrt{7} + 2)(\sqrt{7} - 2)}
\]
Calculating the denominator:
\[
(\sqrt{7} + 2)(\sqrt{7} - 2) = 7 - 4 = 3
\]
Now for the numerator:
\[
(\sqrt{7} - 2)^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}
\]
Putting it all together:
\[
\frac{11 - 4\sqrt{7}}{3}
\]
### 2. \(\frac{3\sqrt{6} - 2}{3\sqrt{6} + 2}\)
The conjugate of the denominator \(3\sqrt{6} + 2\) is \(3\sqrt{6} - 2\).
Now we can multiply the numerator and denominator:
\[
\frac{3\sqrt{6} - 2}{3\sqrt{6} + 2} \cdot \frac{3\sqrt{6} - 2}{3\sqrt{6} - 2} = \frac{(3\sqrt{6} - 2)^2}{(3\sqrt{6} + 2)(3\sqrt{6} - 2)}
\]
Calculating the denominator:
\[
(3\sqrt{6} + 2)(3\sqrt{6} - 2) = (3\sqrt{6})^2 - 2^2 = 54 - 4 = 50
\]
Now for the numerator:
\[
(3\sqrt{6} - 2)^2 = (3\sqrt{6})^2 - 2 \cdot 3\sqrt{6} \cdot 2 + 2^2 = 54 - 12\sqrt{6} + 4 = 58 - 12\sqrt{6}
\]
Putting it all together:
\[
\frac{58 - 12\sqrt{6}}{50}
\]
### Final Answers:
1. \(\frac{11 - 4\sqrt{7}}{3}\)
2. \(\frac{58 - 12\sqrt{6}}{50}\)
### 1. \(\frac{\sqrt{7} - 2}{\sqrt{7} + 2}\)
The conjugate of the denominator \(\sqrt{7} + 2\) is \(\sqrt{7} - 2\).
Now we can multiply the numerator and denominator:
\[
\frac{\sqrt{7} - 2}{\sqrt{7} + 2} \cdot \frac{\sqrt{7} - 2}{\sqrt{7} - 2} = \frac{(\sqrt{7} - 2)^2}{(\sqrt{7} + 2)(\sqrt{7} - 2)}
\]
Calculating the denominator:
\[
(\sqrt{7} + 2)(\sqrt{7} - 2) = 7 - 4 = 3
\]
Now for the numerator:
\[
(\sqrt{7} - 2)^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}
\]
Putting it all together:
\[
\frac{11 - 4\sqrt{7}}{3}
\]
### 2. \(\frac{3\sqrt{6} - 2}{3\sqrt{6} + 2}\)
The conjugate of the denominator \(3\sqrt{6} + 2\) is \(3\sqrt{6} - 2\).
Now we can multiply the numerator and denominator:
\[
\frac{3\sqrt{6} - 2}{3\sqrt{6} + 2} \cdot \frac{3\sqrt{6} - 2}{3\sqrt{6} - 2} = \frac{(3\sqrt{6} - 2)^2}{(3\sqrt{6} + 2)(3\sqrt{6} - 2)}
\]
Calculating the denominator:
\[
(3\sqrt{6} + 2)(3\sqrt{6} - 2) = (3\sqrt{6})^2 - 2^2 = 54 - 4 = 50
\]
Now for the numerator:
\[
(3\sqrt{6} - 2)^2 = (3\sqrt{6})^2 - 2 \cdot 3\sqrt{6} \cdot 2 + 2^2 = 54 - 12\sqrt{6} + 4 = 58 - 12\sqrt{6}
\]
Putting it all together:
\[
\frac{58 - 12\sqrt{6}}{50}
\]
### Final Answers:
1. \(\frac{11 - 4\sqrt{7}}{3}\)
2. \(\frac{58 - 12\sqrt{6}}{50}\)
Answered by
GPT-4o mini
Let's carefully break down the calculation of the denominator for the first surd fraction, which is:
\[
\frac{\sqrt{7} - 2}{\sqrt{7} + 2}
\]
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[
\sqrt{7} + 2 \Rightarrow \text{conjugate: } \sqrt{7} - 2
\]
So we have:
\[
\frac{\sqrt{7} - 2}{\sqrt{7} + 2} \cdot \frac{\sqrt{7} - 2}{\sqrt{7} - 2}
\]
The denominator in this expression becomes:
\[
(\sqrt{7} + 2)(\sqrt{7} - 2)
\]
This is a difference of squares formula, \(a^2 - b^2\), where:
- \(a = \sqrt{7}\)
- \(b = 2\)
Calculating it:
\[
(\sqrt{7})^2 - (2)^2 = 7 - 4 = 3
\]
So the denominator simplifies correctly to:
\[
3
\]
Thus, the correct calculation for the denominator is:
\[
\sqrt{7}^2 - 2^2 = 7 - 4 = 3
\]
The result matches what we found earlier. Thank you for your question, and I'm glad we could clarify this detail!
\[
\frac{\sqrt{7} - 2}{\sqrt{7} + 2}
\]
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[
\sqrt{7} + 2 \Rightarrow \text{conjugate: } \sqrt{7} - 2
\]
So we have:
\[
\frac{\sqrt{7} - 2}{\sqrt{7} + 2} \cdot \frac{\sqrt{7} - 2}{\sqrt{7} - 2}
\]
The denominator in this expression becomes:
\[
(\sqrt{7} + 2)(\sqrt{7} - 2)
\]
This is a difference of squares formula, \(a^2 - b^2\), where:
- \(a = \sqrt{7}\)
- \(b = 2\)
Calculating it:
\[
(\sqrt{7})^2 - (2)^2 = 7 - 4 = 3
\]
So the denominator simplifies correctly to:
\[
3
\]
Thus, the correct calculation for the denominator is:
\[
\sqrt{7}^2 - 2^2 = 7 - 4 = 3
\]
The result matches what we found earlier. Thank you for your question, and I'm glad we could clarify this detail!
Answered by
GPT-4o mini
You're very welcome! I'm glad I could help. If you have any more questions or need further assistance, feel free to ask. Happy studying! π«ΆπΌ
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