Question

If $x=3+\sqrt{8}$, find the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$.

Answer 1

Hint:In this given question, we can use the corollary of the formula of the square of the sum of two numbers in order to find the value of the asked polynomial ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$, that is,
$\begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab \\\ \end{aligned}$.
Then by using $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we will arrive at the answer.

Complete step-by-step answer:
In this given question, we are given that the value of $x$ is equal to $3+\sqrt{8}$, that is $x=3+\sqrt{8}$.
We are asked to find out the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$.
While getting a solution to this question we will be using use the corollary of the formula of the square of the sum of two numbers, that is,
$\begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab................(1.1) \\\ \end{aligned}$
And the formula of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}.............(1.2)$.
The process of solving is as follows:
By replacing $a$ and $b$ with $x$ and $\dfrac{1}{x}$ in equation 1.1, we obtain
$\begin{aligned} & {{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{\left( x+\dfrac{1}{x} \right)}^{2}}-2\times x\times \dfrac{1}{x} \\\ & \Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}={{\left( x+\dfrac{1}{x} \right)}^{2}}-2.............(1.3) \\\ \end{aligned}$
We are given the value of $x$ as equal to $3+\sqrt{8}$.
So, putting this value of $x$ in equation 1.2, we get
${{\left( 3+\sqrt{8} \right)}^{2}}+{{\left( \dfrac{1}{3+\sqrt{8}} \right)}^{2}}={{\left( 3+\sqrt{8}+\dfrac{1}{3+\sqrt{8}} \right)}^{2}}-2$

$={{\left( \dfrac{{{\left( 3+\sqrt{8} \right)}^{2}}+1}{3+\sqrt{8}} \right)}^{2}}-2............(1.4)$
Now, rationalizing the denominator of equation 1.4 by multiplying $\left( 3-\sqrt{8} \right)$in both the denominator and numerator and then using equation 1.2, we get

& {{\left( \dfrac{{{\left( 3+\sqrt{8} \right)}^{2}}+1}{3+\sqrt{8}} \right)}^{2}}-2={{\left( \dfrac{\left[ {{\left( 3+\sqrt{8} \right)}^{2}}+1 \right]\left( 3-\sqrt{8} \right)}{\left( 3+\sqrt{8} \right)\left( 3-\sqrt{8} \right)} \right)}^{2}}-2 \\\ & ={{\left( \dfrac{\left( 3+\sqrt{8} \right)\left( 3+\sqrt{8} \right)\left( 3-\sqrt{8} \right)+1\left( 3-\sqrt{8} \right)}{{{3}^{2}}-{{\left( \sqrt{8} \right)}^{2}}} \right)}^{2}}-2 \\\ & ={{\left( \dfrac{\left( 3+\sqrt{8} \right)\left( {{3}^{2}}-{{\left( \sqrt{8} \right)}^{2}} \right)+\left( 3-\sqrt{8} \right)}{9-8} \right)}^{2}}-2 \\\ & ={{\left( \dfrac{\left( 3+\sqrt{8} \right)\left( 9-8 \right)+\left( 3-\sqrt{8} \right)}{1} \right)}^{2}}-2..............(1.5) \\\ \end{aligned}$$ Therefore, the denominator has been simplified and we are left with the terms in the numerator and thus from equation (1.5), we have $$\begin{aligned} & {{\left( \dfrac{{{\left( 3+\sqrt{8} \right)}^{2}}+1}{3+\sqrt{8}} \right)}^{2}}-2={{\left[ \left( 3+\sqrt{8} \right)+\left( 3-\sqrt{8} \right) \right]}^{2}}-2 \\\ & ={{\left( 3+\sqrt{8}+3-\sqrt{8} \right)}^{2}}-2 \\\ & ={{6}^{2}}-2=36-2=34 \\\ \end{aligned}$$ Hence, we have got the value of the above simplification as 34. Therefore, we get the answer of our question as 34 that is equal to the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$, where $x=3+\sqrt{8}$. Note: In equation (1.5), we should note that we could also have found the value by explicitly calculating the values in the numerator and the denominator. However, as $\sqrt{8}$ is not an integer, the calculation would have been difficult. Therefore, rationalizing the denominator is a better method which we have adopted while solving the problem.