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How do you simplify: square root of 300 divided by the square root of 6?

seo-qna
Last updated date: 20th Jun 2024
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Answer
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Hint:We first explain the process of root finding in case of fractions. The root of the whole fraction happens by finding the root of the denominator and numerator separately as \[\dfrac{\sqrt{p}}{\sqrt{q}}=\sqrt{\dfrac{p}{q}}\]. We find the root values and place to find the solution. We try to form the indices formula for the value 2. We find the prime factorisation of 50.

Complete step by step solution:
We need to find the square root of 300 divided by the square root of 6.
Square root of 300 is defined as $\sqrt{300}$. Square root of 6 is defined as $\sqrt{6}$.
We have to solve the answer of $\dfrac{\sqrt{300}}{\sqrt{6}}$. We know the theorem of indices
\[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt{a}\]. The main part in the root is a fraction.
For a general fraction $\dfrac{p}{q}$, the root of the whole thing works as separately taken roots of denominator and the numerator.
So, \[\dfrac{\sqrt{p}}{\sqrt{q}}=\sqrt{\dfrac{p}{q}}\].
This means we need to find the answer of $\sqrt{\dfrac{300}{6}}$ instead of finding \[\sqrt{300}\]
and $\sqrt{6}$.
Now we find the division where $\dfrac{300}{6}=50$. So, $\sqrt{\dfrac{300}{6}}=\sqrt{50}$
For $\sqrt{50}$, the prime factorisation of 50 will be
$\begin{align}
& 2\left| \!{\underline {\,
50 \,}} \right. \\
& 5\left| \!{\underline {\,
25 \,}} \right. \\
& 5\left| \!{\underline {\,
5 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}$
Therefore, $\sqrt{50}=\sqrt{2\times 5\times 5}=5\sqrt{2}$.
The simplified value of the square root $\dfrac{\sqrt{300}}{\sqrt{6}}$ is $5\sqrt{2}$.

Note: The thing we need to remember is that in normal cases of finding roots we find two values for square roots. But in this case, we only found and used positive results. We didn’t consider the negative part as the sign of the root was already provided as positive. If it was told to find a root of $50$, then we would have taken both positive and negative values.