Question

If $x=1-\sqrt{2}$, find the value of $x-\dfrac{1}{x}$

Answer 1

Hint: We have given expression as $x=1-\sqrt{2}$. Divide 1 by the given expression to get the value of $\dfrac{1}{x}$. Use the rationalization method to rationalize $\dfrac{1}{x}$ and get the value in simple terms. Then, subtract $\dfrac{1}{x}$ from x to get the value of $x-\dfrac{1}{x}$.

Complete step-by-step answer:
Since, we have a given equation as: $x=1-\sqrt{2}......(1)$
Now, we need to find the value of $\dfrac{1}{x}$
So, divide 1 by equation (1), we get:
\[\dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}......(2)\]
Now, we need to rationalize equation (2).
To rationalize a given expression, multiply and divide by same number., i.e.
$\dfrac{1}{a+b}\times \dfrac{a-b}{a-b}$
So, for given equation (2), by rationalizing, we can write:
$\dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}......(3)$
Now, by using the identity: $\left( a-b \right)\left( a+b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$
We can write equation (3) as:
$\begin{align}
  & \dfrac{1}{x}=\dfrac{1+\sqrt{2}}{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & =\dfrac{1+\sqrt{2}}{1-2} \\
 & =-\left( 1+\sqrt{2} \right)......(4)
\end{align}$
Now, we have the value of x and $\dfrac{1}{x}$. We need to find the value of $x-\dfrac{1}{x}$
So, subtract equation (4) from equation (1), we get:
$\begin{align}
  & x-\dfrac{1}{x}=\left( 1-\sqrt{2} \right)-\left( -\left( 1+\sqrt{2} \right) \right) \\
 & =\left( 1-\sqrt{2} \right)+\left( 1+\sqrt{2} \right) \\
 & =2
\end{align}$
Hence, the value of $x-\dfrac{1}{x}$ is 2

Note: We can solve the expression without rationalizing the value of $\dfrac{1}{x}$
As we need to find the value of $x-\dfrac{1}{x}$, subtract equation (2) from equation (1), we get:
$x-\dfrac{1}{x}=\left( 1-\sqrt{2} \right)-\dfrac{1}{\left( 1-\sqrt{2} \right)}$
Now, by taking LCM we can write:
\[x-\dfrac{1}{x}=\dfrac{{{\left( 1-\sqrt{2} \right)}^{2}}-1}{\left( 1-\sqrt{2} \right)}\]
Now, using identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we can write:
\[\begin{align}
  & x-\dfrac{1}{x}=\dfrac{1+2-2\sqrt{2}-1}{\left( 1-\sqrt{2} \right)} \\
 & =\dfrac{2-2\sqrt{2}}{\left( 1-\sqrt{2} \right)}
\end{align}\]
Now take 2 as common from numerator, we get:
\[\begin{align}
  & x-\dfrac{1}{x}=\dfrac{2\left( 1-\sqrt{2} \right)}{\left( 1-\sqrt{2} \right)} \\
 & =2
\end{align}\]
Hence, the value of $x-\dfrac{1}{x}$ is 2
The expression given was simpler. But if we get a complex expression which involves a number in square root, always use the rationalization method. It makes the solution easier. The other method is a bit lengthy. Also, remember while applying the rationalization method, multiply numerator and denomination by the additive inverse of the given expression.