Question

How do you simplify $\sqrt{\dfrac{27}{64}}$?

Answer 1

Assume the given expression as ‘E’. Write the given numerator and the denominator as the product of their prime factors. Cancel the common factors if present. Now, try to write the prime factors with their exponent equal to 2. To do this, form groups of identical prime factors. Now, apply the formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to cancel the square root and exponent 2, to get the answer.

Complete answer:
Here, we have been asked to simplify the expression $\sqrt{\dfrac{27}{64}}$. That means we have to find the square root of the fraction $\dfrac{27}{64}$. Let us assume the given expression as ‘E’. So, we have,
$\Rightarrow E=\sqrt{\dfrac{27}{64}}$
Now, to find the square root, first we need to write the numerator and the denominator as the product of their factors. So, we can write: -

& \Rightarrow 27=3\times 3\times 3 \\\ & \Rightarrow 64=2\times 2\times 2\times 2\times 2\times 2 \\\ \end{aligned}$$ So, the expression (E) becomes: - $$\Rightarrow E=\sqrt{\dfrac{3\times 3\times 3}{2\times 2\times 2\times 2\times 2\times 2}}$$ Since, we have to find the square root so we need to form a group of two identical prime factors so that we can make the exponent equal to 2. So, grouping the identical prime factors, we have, $$\Rightarrow E=\sqrt{\dfrac{\left( 3\times 3 \right)\times 3}{\left( 2\times 2 \right)\times \left( 2\times 2 \right)\times \left( 2\times 2 \right)}}$$ In the exponential form we can write the expression as: - $$\Rightarrow E=\sqrt{\dfrac{{{3}^{2}}\times 3}{{{2}^{2}}\times {{2}^{2}}\times {{2}^{2}}}}$$ Using the formula: - $${{a}^{m}}\times {{b}^{m}}={{\left( a\times b \right)}^{m}}$$, in the denominator, we have, $$\begin{aligned} & \Rightarrow E=\sqrt{\dfrac{{{3}^{2}}\times 3}{{{\left( 2\times 2\times 2 \right)}^{2}}}} \\\ & \Rightarrow E=\sqrt{\dfrac{{{3}^{2}}\times 3}{{{8}^{2}}}} \\\ \end{aligned}$$ Now, in the numerator one of the prime factors 3 is left alone and cannot be paired, so to make exponent equal to 2 we can write $$3={{\left( \sqrt{3} \right)}^{2}}$$, so we get, $$\begin{aligned} & \Rightarrow E=\sqrt{\dfrac{{{3}^{2}}\times {{\left( \sqrt{3} \right)}^{2}}}{{{8}^{2}}}} \\\ & \Rightarrow E=\sqrt{\dfrac{{{\left( 3\sqrt{3} \right)}^{2}}}{{{8}^{2}}}} \\\ \end{aligned}$$ The above expression can be simplified as: - $$\Rightarrow E=\sqrt{{{\left( \dfrac{3\sqrt{3}}{8} \right)}^{2}}}$$ Now, square root means exponent equal to $$\dfrac{1}{2}$$, so we have, $$\Rightarrow E={{\left[ {{\left( \dfrac{3\sqrt{3}}{8} \right)}^{2}} \right]}^{\dfrac{1}{2}}}$$ Using the formula: - $${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$$, we get, $$\begin{aligned} & \Rightarrow E={{\left( \dfrac{3\sqrt{3}}{8} \right)}^{2\times \dfrac{1}{2}}} \\\ & \Rightarrow E=\dfrac{3\sqrt{3}}{8} \\\ \end{aligned}$$ Hence, the simplified form of $$\sqrt{\dfrac{27}{64}}$$ is $$\dfrac{3\sqrt{3}}{8}$$. **Note:** You can find the simplified form in decimal also by using the substitution $$\sqrt{3}=1.732$$. You must know how to find prime factors of a number because in many cases the given number will be very large and in such cases we would have no option other than finding the prime factors. Note that here we were asked to find the square root and that is why we have formed a group of two identical factors. If we were to find a cube root then we would have formed a group of three identical factors.