How do you simplify: square root $\left( \dfrac{3}{5} \right)$?
Answer
621.6k+ views
Hint: In this problem we need to simplify the value of the square root of $\left( \dfrac{3}{5} \right)$. Mathematically it can be written as $\sqrt{\dfrac{3}{5}}$. Now we will apply the exponential rule i.e., $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ and simplify the given value. After that we can observe a square root value in the denominator of the fraction. So, we will rationalize the obtained fraction by multiplying and dividing the obtained fraction with the denominator of that fraction. Now we will apply the exponential formulas $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$, $\sqrt{a}\times \sqrt{a}=a$ and simplify the fraction to get the required result.
Complete step by step answer:
Given that, the square root of $\left( \dfrac{3}{5} \right)$.
Mathematically we can write it as $\sqrt{\dfrac{3}{5}}$.
We can observe that the above value is in the form of $\sqrt{\dfrac{a}{b}}$, we have the formula for the $\sqrt{\dfrac{a}{b}}$ as $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$. Applying this formula to the given value, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{3}}{\sqrt{5}}$
In the above value we can observe that the denominator of the above fraction is $\sqrt{5}$. To rationalize the above fraction, we are going to multiply and divide the above fraction with same $\sqrt{5}$, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{3}}{\sqrt{5}}\times \dfrac{\sqrt{5}}{\sqrt{5}}$
Multiplying the numerator with the numerator and denominator with denominator in the above equation, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{3}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$
Applying the formulas, $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$, $\sqrt{a}\times \sqrt{a}=a$ in the above equation and simplifying the above equation, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{15}}{5}$.
Hence the simplified form of the given value is $\dfrac{\sqrt{15}}{5}$.
Note: We can also simplify the given value in another way. That is, we can multiply and divide the given fraction with $5$ within square root, then we will get
$\begin{align}
& \Rightarrow \sqrt{\dfrac{3}{5}}=\sqrt{\dfrac{3\times 5}{5\times 5}} \\
& \Rightarrow \sqrt{\dfrac{3}{5}}=\sqrt{\dfrac{15}{25}} \\
& \Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{15}}{5} \\
\end{align}$
From both the methods we got the same result.
Complete step by step answer:
Given that, the square root of $\left( \dfrac{3}{5} \right)$.
Mathematically we can write it as $\sqrt{\dfrac{3}{5}}$.
We can observe that the above value is in the form of $\sqrt{\dfrac{a}{b}}$, we have the formula for the $\sqrt{\dfrac{a}{b}}$ as $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$. Applying this formula to the given value, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{3}}{\sqrt{5}}$
In the above value we can observe that the denominator of the above fraction is $\sqrt{5}$. To rationalize the above fraction, we are going to multiply and divide the above fraction with same $\sqrt{5}$, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{3}}{\sqrt{5}}\times \dfrac{\sqrt{5}}{\sqrt{5}}$
Multiplying the numerator with the numerator and denominator with denominator in the above equation, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{3}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$
Applying the formulas, $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$, $\sqrt{a}\times \sqrt{a}=a$ in the above equation and simplifying the above equation, then we will get
$\Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{15}}{5}$.
Hence the simplified form of the given value is $\dfrac{\sqrt{15}}{5}$.
Note: We can also simplify the given value in another way. That is, we can multiply and divide the given fraction with $5$ within square root, then we will get
$\begin{align}
& \Rightarrow \sqrt{\dfrac{3}{5}}=\sqrt{\dfrac{3\times 5}{5\times 5}} \\
& \Rightarrow \sqrt{\dfrac{3}{5}}=\sqrt{\dfrac{15}{25}} \\
& \Rightarrow \sqrt{\dfrac{3}{5}}=\dfrac{\sqrt{15}}{5} \\
\end{align}$
From both the methods we got the same result.
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