Question

Find the value of$\sqrt {\dfrac{{12}}{{75}}} $ by simplification.

Answer 1

Hint:The radical properties include the product property of the radicals and the quotient property of the radicals. They are as follows,
$
\sqrt {x \times y} = \sqrt x \times \sqrt y \\
\sqrt {\dfrac{x}{y}} = \dfrac{{\sqrt x }}{{\sqrt y }} \\
$
Where \[x \geqslant 0,y \geqslant 0\].

The answer is always positive and x and y can't be negative since we won’t get a real answer.

Complete step by step solution:
Using the quotient property of the radicals, we can rewrite the question as
$\sqrt {\dfrac{{12}}{{75}}} = \dfrac{{\sqrt {12} }}{{\sqrt {75} }}$

We now have two options. Either we solve to get the square roots of the numerator and the
denominator individually or we simplify the question further. As finding the square roots of not perfect squares are lengthy, let us simplify it further.

Using product property of the radicals, we can rewrite the question further as,
$\dfrac{{\sqrt {12} }}{{\sqrt {75} }} = \dfrac{{\sqrt {4 \times 3} }}{{\sqrt {25 \times 3} }} =
\dfrac{{\sqrt 4 \times \sqrt 3 }}{{\sqrt {25} \times \sqrt 3 }}$

Both numerator and denominator have $\sqrt 3 $ common,
$\sqrt {\dfrac{{12}}{{75}}} = \dfrac{{\sqrt 4 }}{{\sqrt {25} }}$

4 and 25 are perfect squares of 2 and 5.

Thus, $\sqrt {\dfrac{{12}}{{75}}} = \dfrac{{\sqrt 4 }}{{\sqrt {25} }} = \dfrac{2}{5}$

Alternate Method: We may also simplify the question by rationalizing the denominator. The denominator to be rationalized needs to be multiplied by $\sqrt {75} $and subsequently, the numerator must also be multiplied with $\sqrt {75} $. The rewritten expression looks like,

$\sqrt {\dfrac{{12}}{{75}}} = \dfrac{{\sqrt {12} \times \sqrt {75} }}{{\sqrt {75} \times \sqrt {75} }}
= \dfrac{{\sqrt {12 \times 75} }}{{{{\sqrt {75} }^2}}} = \dfrac{{\sqrt {900} }}{{75}} =
\dfrac{{30}}{{75}}$

This can be further simplified to, $\dfrac{2}{5}$by dividing both the numerator and denominator by 15.

Note: The symbol $\sqrt[n]{x}$ indicates a root and is termed a radical. It is read as “x radical n," or "the nth root of x ". Within the radical symbol, the horizontal line is termed the vinculum. The amount under the vinculum is named the radicand, and the quantity n written to the left is
named the index.