Question

Find the value of \[\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right) \], if ${\text{A = 5 + 2}}\sqrt {\text{6}} $
a) $2\sqrt 3 $
b) $3\sqrt 3 $
c) $2\sqrt 2 $
d) $\sqrt 3 $

Answer 1

Hint: In this question we have to find the required relation value. That relation was formed by the given ${\text{A}}$. They gave the value of ${\text{A}}$. By using the value of ${\text{A}}$ we are going to find the value of the required relation. For that, we are going to solve using an algebra formula and conjugate the values. Algebra covers the simple operation of mathematics like addition, subtraction, multiplication, and division involving both constants as well as variables.

Formulas used:
${\left( {{\text{a + b}}} \right)^{\text{2}}}{\text{ = (}}{{\text{a}}^{\text{2}}}{\text{ + 2ab + }}{{\text{b}}^{\text{2}}}{\text{)}}$

Complete step by step answer:
It is given that the value of ${\text{A}}$ is ${\text{A = 5 + 2}}\sqrt {\text{6}} $,
If ${\text{A = x}}$, then
\[{\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)^2}\]
will be using the formula already mentioned
${\left( {{\text{a + b}}} \right)^{\text{2}}}{\text{ = (}}{{\text{a}}^{\text{2}}}{\text{ + 2ab + }}{{\text{b}}^{\text{2}}}{\text{)}}$,
Now,
\[{\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)^2} = {\left( {\sqrt x } \right)^2} + 2\sqrt x \dfrac{1}{{\sqrt x }} + {\left( {\dfrac{1}{{\sqrt x }}} \right)^2}\]
Solving the square root by squaring the roots,
$ \Rightarrow {\text{x + 2 + }}\dfrac{{\text{1}}}{{\text{x}}}$
In our problem the given of ${\text{A}}$ is ${\text{A = 5 + 2}}\sqrt {\text{6}} $,
Substituting the value of ${\text{A}}$ we get,
$\Rightarrow {\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)^2} = 5 + 2\sqrt 6 + 2 + \dfrac{1}{{5 + 2\sqrt 6 }} - - - - - - - \left( 1 \right)$
To solve the equation, we have to conjugate the terms,
First we take conjugate for $\dfrac{1}{{5 + 2\sqrt 6 }}$ and substitute the value in equation 1
$ \Rightarrow \dfrac{1}{{5 + 2\sqrt 6 }} \times \dfrac{{5 - 2\sqrt 6 }}{{5 - 2\sqrt 6 }}$
Multiplying the terms we get,
$ \Rightarrow \dfrac{{5 - 2\sqrt 6 }}{{{{\left( 5 \right)}^2} - {{\left( {2\sqrt 6 } \right)}^2}}}$
Squaring the denominator to remove the square root,
$ \Rightarrow \dfrac{{5 - 2\sqrt 6 }}{{25 - 4\left( 6 \right)}}$
Multiplying the terms in denominator we get,
$ \Rightarrow \dfrac{{5 - 2\sqrt 6 }}{{25 - 24}}$
Subtracting the terms we get,
$ \Rightarrow \dfrac{{5 - 2\sqrt 6 }}{1}$
Hence we get,
$\Rightarrow \dfrac{1}{{5 + 2\sqrt 6 }} = 5 - 2\sqrt 6 $
Substituting the terms in equation (1),
$\Rightarrow {\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)^2} = 5 + 2\sqrt 6 + 2 + \dfrac{1}{{5 + 2\sqrt 6 }}$
It will becomes,
$ \Rightarrow 5 + 2\sqrt 6 + 2 + 5 - 2\sqrt 6 $
Simplifying the terms we get,
$\Rightarrow {\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)^2} = 12$
In our problem we need to find the value of \[\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)\],
We found the term ${\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right)^2} = 12$, just taking square root in this term will give our required term. Hence we have to take a square root on both sides,
$\Rightarrow \left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right) = \sqrt {12} $
Since we know that,
$\Rightarrow \sqrt {12} = \sqrt {2 \times 2 \times 3} = 2\sqrt 3 $

$\therefore $ The value of $\left( {\sqrt {\text{A}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{A}} }}} \right) = 2\sqrt 3 $

Note:
In this problem, we found the value of the required term by using the given term. We substituted the given term into some algebraic relations and formulas to find the required term. There is some simple calculation only in the solution, but students may go wrong with careless mistakes. So students must have concentrated on these calculations.