Simplify Square Root Expression $\sqrt{125} - \sqrt{5}$ $\sqrt{125}-\sqrt5$ simplify it.
I thought it would be $\sqrt {5\cdot5\cdot5}-\sqrt 5$ which would be the square root of 25 which is 5 but it is not.
Can you show how to simplify this?
 A: $\sqrt{125}-\sqrt{5}=\sqrt{5\times 25}-\sqrt{5}=\sqrt{25}\sqrt{5}-\sqrt{5}=5\sqrt{5}-\sqrt{5}=4\sqrt{5}$
A: $$(\sqrt{125}-\sqrt5)^2=125+5-2\sqrt{625}=130-50=80$$
therefore
$$\sqrt{125}-\sqrt 5=\sqrt{80}=4\sqrt 5$$
It is not the most usual way to do this. Just giving an alternative solution.
A: So you are asking 
$$ \sqrt{125} - \sqrt{5} $$
This is
$$ \sqrt{25 \times 5} - \sqrt{5} $$ 
Which simplifies to 
$$ 5 \sqrt{5} - \sqrt{5} $$ 
Which factors to 
$$ \sqrt{5}(5 - 1) $$ 
Thus the answer is 
$$ 4 \sqrt{5} $$ 
A: Try thinking about it as $ax - bx$. The most obvious choice is $x = \sqrt{5}$, so that then $b = 1$. Now you just need to rewrite $\sqrt{125}$ as $a \sqrt{5}$, and it turns out that $\sqrt{125} = 5 \sqrt{5}$.
So now you just do $5 \sqrt{5} - \sqrt{5} = 4 \sqrt{5}$.
A: Starting from where you left off:
$$
\begin{eqnarray}
\sqrt{125}-\sqrt5
&=& \sqrt {5\cdot5\cdot5} - \sqrt 5 \\
&=& \sqrt{5} \cdot \sqrt{5} \cdot\sqrt{5}-\sqrt 5  \\
&=& \sqrt{5} (\sqrt{5} \cdot\sqrt{5} - 1)  \\
&=& \sqrt{5} (5 - 1)  \\
&=& 4 \sqrt{5}
\end{eqnarray}
$$

If the question were $\sqrt{125} \div \sqrt5$, then the answer would indeed be $5$.
A: \begin{align}
\sqrt{125}-\sqrt{5}&=\sqrt{5^3}-\sqrt{5}\\
                   &=\sqrt{5^2\cdot 5}-\sqrt{5}\\
                   &=\sqrt{5^2}\sqrt{5}-\sqrt{5}\\
                   &=5\sqrt{5}-\sqrt{5}\\
                   &=4\sqrt{5}
\end{align}
