How do you simplify $\sqrt{\dfrac{27}{16}}$?
Answer
618.3k+ views
Hint: First separate the square root for the numerator and the denominator using the formula $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$. Then express both the numerator and the denominator in terms of their prime factors. For the square root, take each pair of factors only once.
Complete step by step answer:
The expression we have $\sqrt{\dfrac{27}{16}}$
As we know $\sqrt{\dfrac{a}{b}}$ can be written as $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
So, $\sqrt{\dfrac{27}{16}}$ can be written as $\sqrt{\dfrac{27}{16}}=\dfrac{\sqrt{27}}{\sqrt{16}}$
To find the square root of a number we have to express it in terms of it’s prime factors. Then we have to take once from each pair of factors.
The prime factors of $27=3\times 3\times 3$
The prime factors of $16=2\times 2\times 2\times 2$
Now, expressing the numerator and the denominator of our expression as their prime factors, we get
$\dfrac{\sqrt{27}}{\sqrt{16}}=\dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}}$
Taking each pair of factors once for the square root, we get
$\begin{align}
& \dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}} \\
& =\dfrac{3\times \sqrt{3}}{2\times 2} \\
& =\dfrac{3\sqrt{3}}{4} \\
\end{align}$
This is the required solution of the given question.
Note:
Separation of the square root function should be done for the numerator and the denominator for the further simplification. For the square root of$\dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}}$, since there is one pair of ‘3’ in the numerator and two pairs of ‘2’ in the denominator so we should take ‘3’ only once in the numerator and ‘2’ twice in the denominator of the result, which can be written as $\dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}}=\dfrac{3\times \sqrt{3}}{2\times 2}=\dfrac{3\sqrt{3}}{4}$.
Complete step by step answer:
The expression we have $\sqrt{\dfrac{27}{16}}$
As we know $\sqrt{\dfrac{a}{b}}$ can be written as $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$
So, $\sqrt{\dfrac{27}{16}}$ can be written as $\sqrt{\dfrac{27}{16}}=\dfrac{\sqrt{27}}{\sqrt{16}}$
To find the square root of a number we have to express it in terms of it’s prime factors. Then we have to take once from each pair of factors.
The prime factors of $27=3\times 3\times 3$
The prime factors of $16=2\times 2\times 2\times 2$
Now, expressing the numerator and the denominator of our expression as their prime factors, we get
$\dfrac{\sqrt{27}}{\sqrt{16}}=\dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}}$
Taking each pair of factors once for the square root, we get
$\begin{align}
& \dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}} \\
& =\dfrac{3\times \sqrt{3}}{2\times 2} \\
& =\dfrac{3\sqrt{3}}{4} \\
\end{align}$
This is the required solution of the given question.
Note:
Separation of the square root function should be done for the numerator and the denominator for the further simplification. For the square root of$\dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}}$, since there is one pair of ‘3’ in the numerator and two pairs of ‘2’ in the denominator so we should take ‘3’ only once in the numerator and ‘2’ twice in the denominator of the result, which can be written as $\dfrac{\sqrt{3\times 3\times 3}}{\sqrt{2\times 2\times 2\times 2}}=\dfrac{3\times \sqrt{3}}{2\times 2}=\dfrac{3\sqrt{3}}{4}$.
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