Question

Evaluate $\dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - \sqrt {3 - 2\sqrt 2 } $

Answer 1

Hint: First break the expression into two parts. Then take the first part and square the term and do simplification. When it is simplified, take the square root of the value. Now take the second part and square the term. After that, break the term in such a way that it forms the square. Then take the square root of the value. In the end, add both terms and do calculations to get the final answer.

Complete step-by-step solution:
Let $x = \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }}$ and $y = \sqrt {3 - 2\sqrt 2 } $.
Then, the term can be written as,
$ \Rightarrow \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - \sqrt {3 - 2\sqrt 2 } = x - y$...................….. (1)
Now take the first term.
$ \Rightarrow x = \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }}$
Square the term,
$ \Rightarrow {x^2} = {\left( {\dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }}} \right)^2}$
We know that,
${\left( {\dfrac{a}{b}} \right)^2} = \dfrac{{{a^2}}}{{{b^2}}}$
Use the identities in the above expression,
$ \Rightarrow {x^2} = \dfrac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}}$
As we know,
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Using this identity, we get,
$ \Rightarrow {x^2} = \dfrac{{{{\left( {\sqrt {\sqrt 5 + 2} } \right)}^2} + {{\left( {\sqrt {\sqrt 5 - 2} } \right)}^2} + 2\left( {\sqrt {\sqrt 5 + 2} } \right)\left( {\sqrt {\sqrt 5 - 2} } \right)}}{{\sqrt 5 + 1}}$
Simplify the term,
$ \Rightarrow {x^2} = \dfrac{{\sqrt 5 + 2 + \sqrt 5 - 2 + 2\left( {\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {2^2}} } \right)}}{{\sqrt 5 + 1}}$
Again, simplify the terms,
$ \Rightarrow {x^2} = \dfrac{{2\sqrt 5 + 2\left( {\sqrt {5 - 4} } \right)}}{{\sqrt 5 + 1}}$
Subtract the value in the bracket,
$ \Rightarrow {x^2} = \dfrac{{2\sqrt 5 + 2\left( {\sqrt 1 } \right)}}{{\sqrt 5 + 1}}$
Take 2 commons in the numerator,
$ \Rightarrow {x^2} = \dfrac{{2\left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}}$
Cancel out the common factors,
$ \Rightarrow {x^2} = 2$
Take the square root on both sides,
$ \Rightarrow x = \sqrt 2 $..................….. (2)
So, the simplified form of the first term is $\sqrt 2 $.
Now take the second term,
$ \Rightarrow y = \sqrt {3 - 2\sqrt 2 } $
Squaring both sides, we get,
$ \Rightarrow {y^2} = {\left( {\sqrt {3 - 2\sqrt 2 } } \right)^2}$
Cancel out square root with square,
$ \Rightarrow {y^2} = 3 - 2\sqrt 2 $
Now break 3 as 1 + 2,
$ \Rightarrow {y^2} = 2 + 1 - 2\sqrt 2 $
Now write 2 as the square of the $\sqrt 2 $ and 1 as the square of 1,
$ \Rightarrow {y^2} = {\left( {\sqrt 2 } \right)^2} + {1^2} - 2\sqrt 2 $
We know that,
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Using the identity, we can write as,
$ \Rightarrow {y^2} = {\left( {\sqrt 2 - 1} \right)^2}$
Take the square root on both sides,
$ \Rightarrow y = \sqrt 2 - 1$.....................….. (3)
So, the simplified form of the second term is $\left( {\sqrt 2 - 1} \right)$,
Substitute the values from equation (2) and (3) in equation (1),
$ \Rightarrow \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - \sqrt {3 - 2\sqrt 2 } = \sqrt 2 - \left( {\sqrt 2 - 1} \right)$
Open bracket and change sign accordingly,
$ \Rightarrow \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - \sqrt {3 - 2\sqrt 2 } = \sqrt 2 - \sqrt 2 + 1$
Simplify the terms,
$\therefore \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - \sqrt {3 - 2\sqrt 2 } = 1$

Hence, the value of $\dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} - \sqrt {3 - 2\sqrt 2 } $ is 1.

Note: You should have an idea of BODMAS which is used to simplify the mathematical expression involving different mathematical operators. The different operators are used according to the rule of BODMAS where,
B= Bracket
O= of
D= Division
M= Multiplication
A = addition
S= subtraction