Question

Find the square root of the algebraic expression which is then average of the following expressions:
${x^2} + \dfrac{1}{{{x^2}}},{\text{ }} - 2\left( {x - \dfrac{1}{x}} \right)$ and $ - 1$.

Answer 1

Hint: First take the average of the given three expressions i.e. ${x^2} + \dfrac{1}{{{x^2}}},{\text{ }} - 2\left( {x - \dfrac{1}{x}} \right)$ and $ - 1$. Reduce the expression obtained in the simplest form and then take its square root as it is asked in the question.

Complete step-by-step answer:
According to the question, the given three expressions are ${x^2} + \dfrac{1}{{{x^2}}},{\text{ }} - 2\left( {x - \dfrac{1}{x}} \right)$ and $ - 1$.We have to find the square root of their average.
So first we will determine the average of three expressions. As we know that the average of any number of observations is determined by the formula:
$ \Rightarrow {\text{Average}} = \dfrac{{{\text{Sum of Observations}}}}{{{\text{Number of observations}}}}$
Applying this formula for the above three expressions, we’ll get:$ \Rightarrow {\text{Average}} = \dfrac{{{x^2} + \dfrac{1}{{{x^2}}} + 2\left( {x - \dfrac{1}{x}} \right) - 1}}{3}$
On adding and subtracting 1 in the numerator, we’ll get:
$ \Rightarrow {\text{Average}} = \dfrac{{{x^2} + \dfrac{1}{{{x^2}}} - 2 + 2\left( {x - \dfrac{1}{x}} \right) + 1}}{3}$
Now, we know that ${a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$. Using this we can conclude that:
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} - 2 = {\left( {x - \dfrac{1}{x}} \right)^2}$
Putting this in above average expression, we’ll get:
$ \Rightarrow {\text{Average}} = \dfrac{{{{\left( {x - \dfrac{1}{x}} \right)}^2} + 2\left( {x - \dfrac{1}{x}} \right) + 1}}{3}$
Again we know that ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$. Using this for numerator, we have:
$ \Rightarrow {\left( {x - \dfrac{1}{x}} \right)^2} + 2\left( {x - \dfrac{1}{x}} \right) + 1 = {\left( {x - \dfrac{1}{x} + 1} \right)^2}$
Putting this in average, we’ll get:
$ \Rightarrow {\text{Average}} = \dfrac{{{{\left( {x - \dfrac{1}{x} + 1} \right)}^2}}}{3}$
Now, we have to take the square root of average. Doing so, we’ll get:
$ \Rightarrow \sqrt {{\text{Average}}} = \sqrt {\dfrac{{{{\left( {x - \dfrac{1}{x} + 1} \right)}^2}}}{3}} $
Simplifying it further, we’ll get:
$
   \Rightarrow \sqrt {{\text{Average}}} = \dfrac{{\sqrt {{{\left( {x - \dfrac{1}{x} + 1} \right)}^2}} }}{{\sqrt 3 }} \\
   \Rightarrow \sqrt {{\text{Average}}} = \dfrac{1}{{\sqrt 3 }}\left| {x - \dfrac{1}{x} + 1} \right| \\
 $

Therefore the square root of the average of the given expressions is $\dfrac{1}{{\sqrt 3 }}\left| {x - \dfrac{1}{x} + 1} \right|$

Note: The square root of any number is always positive. That’s why while taking the square root of any variable we always use modulus after it. The modulus will make it always positive even if the term under square root is negative also. For example,
\[ \Rightarrow {x^2} = 16\]
While taking square root on both sides, the left hand side will come out with modulus because this is the square of a variable.
\[
   \Rightarrow \left| x \right| = 4 \\
   \Rightarrow x = \pm 4 \\
 \]