Question

Find the square roots of \[9 + 2\sqrt {14} \].

Answer 1

Hint: Square root of any number containing unlike terms must have, unlike terms, using this statement we’ll assume a number containing, unlike terms as the root of a given number. Then simplifying that equation to solve for the terms of the square root. So finding those values we’ll find the value of the square root asked.

Complete step by step solution: Given data: The term of which the square root to be found \[ = 9 + 2\sqrt {14} \]
If the square of any number contains unlike terms the number must contain also unlike terms as it can’t be simply further
So now according to the above statement, we can say that the square root of \[9 + 2\sqrt {14} \]contains two unlike terms let say the square root of \[9 + 2\sqrt {14} \] is $(\sqrt a + \sqrt b )$
Therefore according to the above statement, we can say that
\[ \Rightarrow \sqrt {9 + 2\sqrt {14} } = \sqrt a + \sqrt b \]
Squaring both sides and then using \[{\left( {\sqrt a } \right)^2} = a\]
\[ \Rightarrow 9 + 2\sqrt {14} = {\left( {\sqrt a + \sqrt b } \right)^2}\]
On simplifying by using \[{\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}\], we get,
\[ \Rightarrow 9 + 2\sqrt {14} = a + 2\sqrt {ab} + b\]
\[ \Rightarrow 9 + 2\sqrt {14} = (a + b) + 2\sqrt {ab} \]
On comparing the left-hand side and right-hand side
We get, \[9 = a + b\]and \[2\sqrt {ab} = 2\sqrt {14} \]
Now solving \[9 = a + b\]
\[9 - a = b...........(i)\]
Substituting the value of b in \[2\sqrt {ab} = 2\sqrt {14} \] we get
\[ \Rightarrow 2\sqrt {(9 - a)a} = 2\sqrt {14} \]
Dividing both sides by 2
\[ \Rightarrow \sqrt {(9 - a)a} = \sqrt {14} \]
Squaring both sides
\[ \Rightarrow a(9 - a) = 14\]
\[ \Rightarrow 9a - {a^2} = 14\]
\[ \Rightarrow {a^2} - 9a + 14 = 0\]
Now we’ll solve it for ‘a’ by splitting the coefficient of ‘a’ in a way that they are the factors of products of the other two coefficients
\[ \Rightarrow {a^2} - (7 + 2)a + 14 = 0\]
Simplifying the brackets
\[ \Rightarrow {a^2} - 7a - 2a + 14 = 0\]
\[ \Rightarrow a(a - 7) - 2(a - 7) = 0\]
Taking (a-7) common
\[ \Rightarrow (a - 7)(a - 2) = 0\]
i.e. \[a - 7 = 0\]or \[a - 2 = 0\]
\[\therefore a = 7\]or \[a = 2\]
Substituting the value of ‘a’ in equation(i)
When a=7 then b=2
And when a=2 then b=7
From both the case we get the square root as $(\sqrt 7 + \sqrt 2 )$as $a + b = b + a$
But square roots always exist in pair that if ${x^2} = {a^2}$ then $x = \pm a$

Therefore the square root of \[9 + 2\sqrt {14} \] is $ \pm (\sqrt 7 + \sqrt 2 )$

Note: Finding square root means we need to find such a number when multiplied by itself gives that number for which we need to find square root. There are many methods to find square roots. for example, repeated subtraction, prime factorisation, division method and by estimating the square root.