Question
Subtract the radical expressions. \[ s \sqrt{48}-4 \sqrt{27 s^{2}}= \]
Ask by Daniels Sandoval. in the United States
Dec 02,2024
Real Tutor Solution
Tutor-Verified Answer
Answer
\(-8s\sqrt{3}\)
Solution
To subtract the radical expressions \( s \sqrt{48} - 4 \sqrt{27 s^2} \), we first simplify each term.
1. **Simplifying \( s \sqrt{48} \)**:
\[
\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}
\]
Therefore,
\[
s \sqrt{48} = s \cdot 4\sqrt{3} = 4s\sqrt{3}
\]
2. **Simplifying \( 4 \sqrt{27 s^2} \)**:
\[
\sqrt{27 s^2} = \sqrt{9 \cdot 3 \cdot s^2} = \sqrt{9} \cdot \sqrt{3} \cdot \sqrt{s^2} = 3\sqrt{3} \cdot s = 3s\sqrt{3}
\]
Therefore,
\[
4 \sqrt{27 s^2} = 4 \cdot 3s\sqrt{3} = 12s\sqrt{3}
\]
Now we can substitute these simplified forms back into the original expression:
\[
s \sqrt{48} - 4 \sqrt{27 s^2} = 4s\sqrt{3} - 12s\sqrt{3}
\]
Next, we combine the like terms:
\[
4s\sqrt{3} - 12s\sqrt{3} = (4 - 12)s\sqrt{3} = -8s\sqrt{3}
\]
Thus, the final result is:
\[
\boxed{-8s\sqrt{3}}
\]
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Mind Expander
First, let's simplify the radical expressions. The term \( s \sqrt{48} \) can be simplified by noticing that \( \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \). Therefore, \( s \sqrt{48} = 4s \sqrt{3} \). Next, for the term \( 4 \sqrt{27 s^{2}} \), we can simplify it as follows: \( \sqrt{27 s^{2}} = \sqrt{9 \cdot 3} \cdot s = 3s \sqrt{3} \), so \( 4 \sqrt{27 s^{2}} = 4 \cdot 3s \sqrt{3} = 12s \sqrt{3} \). Putting this all together, we have: \[ s \sqrt{48} - 4 \sqrt{27 s^{2}} = 4s \sqrt{3} - 12s \sqrt{3} = (4s - 12s) \sqrt{3} = -8s \sqrt{3}. \] Therefore, the final result is: \[ -8s \sqrt{3}. \]
