Suppose the surd is \(a + \sqrt b \) and its square root is \(\sqrt x  + \sqrt y \),  then 

\(\sqrt {a + \sqrt b }  = \sqrt x  + \sqrt y \)

Squaring both the sides,

\(a + \sqrt b  = x + y + 2\sqrt {xy} \) ………. (1)

Comparing rational and irrational parts, we have, 

\(x + y = a\)  and  \(4xy = b\)

Solving the above two equations we get the values of \(x\) and \(y\).

In most of the questions, we can obtain square root of a surd without solving the above equation. Let us write the equation (1) again,

\[a + \sqrt b  = \underbrace {x + y}_{{\rm{rational}}\;{\rm{part}}} + \underbrace {2\sqrt {xy} }_{{\rm{irrational}}\;{\rm{part}}}\]

Suppose \(x + y = p\) and \(xy = q\), then 

\[a + \sqrt b  = p + 2\sqrt q \]

Now factorize \(q\) into two factors such that their sum is \(p\). If these factors are \(x\) and \(y\), then the square roots is \(\sqrt x  + \sqrt y \). For example,

\(\sqrt {8 + 2\sqrt {15} }  = \sqrt {8 + 2\sqrt 5 \sqrt 3 }  = \sqrt {5 + 3 + 2\sqrt 5 \sqrt 3 } \) =\(\sqrt 5  + \sqrt 3 \).

Here two factors of 15 are 5 and 3 such that their sum is 8.

Now consider the surd,

\({\left( {\sqrt a  + \sqrt b  + \sqrt c } \right)^2} = a + b + c + 2\left( {\sqrt {ab}  + \sqrt {bc}  + \sqrt {ca} } \right)\) 

\(\Rightarrow \sqrt {a + b + c + 2\left( {\sqrt {ab}  + \sqrt {bc}  + \sqrt {ca} } \right)}  = \sqrt a  + \sqrt b  + \sqrt c \)

In such cases, we can factorize the terms \(ab,\;bc\) and \(ca\) into two factors each such that sum of the factors is \(a + b + c\).  

Let us take an example,

\(\sqrt {10 + 2\sqrt 6  + 2\sqrt {10}  + 2\sqrt {15} } \), we have to factorize 6, 10 and 15 into two factors such that sum of these factors is 10. So the factors are (2, 3), (2, 5) and (3, 5) and 2 + 3 + 5 = 10.

Thus \(\sqrt {10 + 2\sqrt 6  + 2\sqrt {10}  + 2\sqrt {15} } \) can be written as 

\(\sqrt {3 + 2 + 5 + 2\sqrt {2.3}  + 2\sqrt {5.2}  + 2\sqrt {3.5} }  = \sqrt 2  + \sqrt 3  + \sqrt 5 \)

Example 01: Find the square root of \(6 - \sqrt 6  - \sqrt {14}  + \sqrt {21} \)