Question
show me a square root simble
Answers
GPT-4o mini
The square root symbol is represented as \( \sqrt{} \). For example, the square root of 16 is written as \( \sqrt{16} \), which equals 4.
Simplify. Rationalize the denominator
-square root7 / 3 + square root 3
-square root7 / 3 + square root 3
GPT-4o mini
To simplify and rationalize the denominator of the expression \(\frac{\sqrt{7}}{3 + \sqrt{3}}\), we can multiply the numerator and denominator by the conjugate of the denominator, which is \(3 - \sqrt{3}\).
Here are the steps:
1. **Multiply by the conjugate:**
\[
\frac{\sqrt{7}}{3 + \sqrt{3}} \cdot \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{\sqrt{7}(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})}
\]
2. **Simplify the denominator:**
The denominator is a difference of squares:
\[
(3 + \sqrt{3})(3 - \sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6
\]
3. **Expand the numerator:**
\[
\sqrt{7}(3 - \sqrt{3}) = 3\sqrt{7} - \sqrt{21}
\]
4. **Combine the results:**
Now we have:
\[
\frac{3\sqrt{7} - \sqrt{21}}{6}
\]
5. **Final simplification:**
We can separate the terms in the fraction:
\[
\frac{3\sqrt{7}}{6} - \frac{\sqrt{21}}{6} = \frac{\sqrt{7}}{2} - \frac{\sqrt{21}}{6}
\]
Thus, the final simplified and rationalized expression is:
\[
\frac{\sqrt{7}}{2} - \frac{\sqrt{21}}{6}
\]
Here are the steps:
1. **Multiply by the conjugate:**
\[
\frac{\sqrt{7}}{3 + \sqrt{3}} \cdot \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{\sqrt{7}(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})}
\]
2. **Simplify the denominator:**
The denominator is a difference of squares:
\[
(3 + \sqrt{3})(3 - \sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6
\]
3. **Expand the numerator:**
\[
\sqrt{7}(3 - \sqrt{3}) = 3\sqrt{7} - \sqrt{21}
\]
4. **Combine the results:**
Now we have:
\[
\frac{3\sqrt{7} - \sqrt{21}}{6}
\]
5. **Final simplification:**
We can separate the terms in the fraction:
\[
\frac{3\sqrt{7}}{6} - \frac{\sqrt{21}}{6} = \frac{\sqrt{7}}{2} - \frac{\sqrt{21}}{6}
\]
Thus, the final simplified and rationalized expression is:
\[
\frac{\sqrt{7}}{2} - \frac{\sqrt{21}}{6}
\]