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?
$\endgroup$6 Answers
$\begingroup$$\sqrt{125}-\sqrt{5}=\sqrt{5\times 25}-\sqrt{5}=\sqrt{25}\sqrt{5}-\sqrt{5}=5\sqrt{5}-\sqrt{5}=4\sqrt{5}$
$\endgroup$ $\begingroup$$$(\sqrt{125}-\sqrt5)^2=125+5-2\sqrt{625}=130-50=80$$ therefore $$\sqrt{125}-\sqrt 5=\sqrt{80}=4\sqrt 5$$
It is not the most usual way to do this. Just giving an alternative solution.
$\endgroup$ 2 $\begingroup$So you are asking
$$ \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} $$
$\endgroup$ $\begingroup$Try thinking about it as $ax - bx$. The most obvious choice is $x = \sqrt{5}$, so that then $b = 1$. Now you just need to rewrite $\sqrt{125}$ as $a \sqrt{5}$, and it turns out that $\sqrt{125} = 5 \sqrt{5}$.
So now you just do $5 \sqrt{5} - \sqrt{5} = 4 \sqrt{5}$.
$\endgroup$ $\begingroup$Starting from where you left off:
$$ \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 $5$.
$\endgroup$ $\begingroup$\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}
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