Question

If we have a variable as $x=1-\sqrt{2}$, then find the value of (i) $x-\dfrac{1}{x}$ and (ii) ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$?

Answer 1

Hint: We start solving the problem by denominator to the term which doesn’t have any fraction to get all terms in a single fraction. We then use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to proceed through the problem and make subsequent arrangements to get the required value of $x-\dfrac{1}{x}$. We then square on both sides and use the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to proceed through the problem. We then make subsequent arrangements and calculations to get the required value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$.

Complete step-by-step solution:
According to the problem, we have given that the value of x is $1-\sqrt{2}$ and we need to find the value of $x-\dfrac{1}{x}$ and ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$.
Let us consider $x-\dfrac{1}{x}$.
$\Rightarrow x-\dfrac{1}{x}=\left( 1-\sqrt{2} \right)-\dfrac{1}{\left( 1-\sqrt{2} \right)}$.
$\Rightarrow x-\dfrac{1}{x}=\dfrac{{{\left( 1-\sqrt{2} \right)}^{2}}-1}{\left( 1-\sqrt{2} \right)}$ --------(1).
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. We use this result in equation (1).
$\Rightarrow x-\dfrac{1}{x}=\dfrac{{{1}^{2}}+{{\left( \sqrt{2} \right)}^{2}}-2\left( 1 \right)\left( \sqrt{2} \right)-1}{\left( 1-\sqrt{2} \right)}$.
$\Rightarrow x-\dfrac{1}{x}=\dfrac{1+2-2\sqrt{2}-1}{\left( 1-\sqrt{2} \right)}$.
$\Rightarrow x-\dfrac{1}{x}=\dfrac{2-2\sqrt{2}}{\left( 1-\sqrt{2} \right)}$ ------(2).
Let us take the number 2 common from both the terms in the numerator.
$\Rightarrow x-\dfrac{1}{x}=\dfrac{2\left( 1-\sqrt{2} \right)}{\left( 1-\sqrt{2} \right)}$.
$\Rightarrow x-\dfrac{1}{x}=2$ ------(3).
We have found the value of $x-\dfrac{1}{x}$ as 2.
Let us do square on both sides in equation (3).
$\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}={{2}^{2}}$ -------(4).
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. We use this result in equation (4).
$\Rightarrow {{x}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}-2\left( x \right)\left( \dfrac{1}{x} \right)=4$.
$\Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2=4$.
$\Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=4+2$.
$\Rightarrow {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=6$.
We have found the value of ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ as 6.
∴ The values of $x-\dfrac{1}{x}$ and ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}$ are 2 and 6.


Note: We need to make sure that we are making calculations related to addition, subtraction, squaring without any mistakes. We can also calculate the value of $\dfrac{1}{x}$ by rationalizing the denominator as the given value of x is an irrational number, we rationalize it by multiplying the given number with its conjugate. The process of finding the value of $\dfrac{1}{x}$ is as shown below:
$\Rightarrow \dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}$.
We know that conjugate of an irrational number $1+\sqrt{r}$ is $1-\sqrt{r}$. So, we get the conjugate of $1-\sqrt{2}$as $1+\sqrt{2}$. We multiply $1+\sqrt{2}$ in both numerator and denominator.
$\Rightarrow \dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}$.
$\Rightarrow \dfrac{1}{x}=\dfrac{1+\sqrt{2}}{{{1}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}$.
$\Rightarrow \dfrac{1}{x}=\dfrac{1+\sqrt{2}}{1-2}$.
$\Rightarrow \dfrac{1}{x}=\dfrac{1+\sqrt{2}}{-1}$.
$\Rightarrow \dfrac{1}{x}=-1-\sqrt{2}$.
We can use this value and calculate the remaining required values.