Answer
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Hint: The trick to solve this question is to take$\sqrt{x}-\dfrac{1}{\sqrt{x}}$as “t” then does the squaring of both the sides. After squaring, the x expression will look as $x+\dfrac{1}{x}-2$, now it is looking simpler as no$\sqrt{x}$ term is here then rationalize the $\dfrac{1}{x}$ expression and then simplify.
Complete step-by-step answer:
Let us take $\sqrt{x}-\dfrac{1}{\sqrt{x}}$as t.
$\sqrt{x}-\dfrac{1}{\sqrt{x}}=t$
Squaring both the sides will change the above equation as follows:
${{\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)}^{2}}={{t}^{2}}$
Now, we are using${{\left( a-b \right)}^{2}}$identity in the above equation and we get,
$x+\dfrac{1}{x}-2={{t}^{2}}$
We are going to substitute the value of x which is equal to 3 + 2$\sqrt{2}$ in the above equation.$3+2\sqrt{2}+\dfrac{1}{3+2\sqrt{2}}-2={{t}^{2}}$
Rationalize the$\dfrac{1}{3+2\sqrt{2}}$. Rationalization is done by multiplying and dividing this rational number by the conjugate of 3 + 2$\sqrt{2}$ and conjugate of 3 + 2$\sqrt{2}$ is 3 -2$\sqrt{2}$. So, the rationalization will resolve the above expression as follows.
$3+2\sqrt{2}+\dfrac{3-2\sqrt{2}}{\left( 3+2\sqrt{2} \right)\left( 3-2\sqrt{2} \right)}-2={{t}^{2}}$
The expression in the denominator is $\left( a+b \right)\left( a-b \right)$ identity which is equal to ${{a}^{2}}-{{b}^{2}}$. So, using this identity in the above expression will give:
$\begin{align}
& 3+2\sqrt{2}+\dfrac{3-2\sqrt{2}}{{{\left( 3 \right)}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}}-2={{t}^{2}} \\
& \Rightarrow 3+2\sqrt{2}+\dfrac{3-2\sqrt{2}}{9-8}-2={{t}^{2}} \\
& \\
\end{align}$
$\Rightarrow 3+2\sqrt{2}+3-2\sqrt{2}-2={{t}^{2}}$
In the above equation, 2$\sqrt{2}$ and - 2$\sqrt{2}$ is cancelled out and the above expression will look as below:
$\begin{align}
& 6-2={{t}^{2}} \\
& \Rightarrow 4={{t}^{2}} \\
& \Rightarrow \pm 2=t \\
\end{align}$
From the above calculation, we have found the value of the expression $\sqrt{x}-\dfrac{1}{\sqrt{x}}$ as ±2.
Hence, the correct option is (b)
Note: If we just substitute the value of x in the expression $\sqrt{x}-\dfrac{1}{\sqrt{x}}$ it will look very complex to solve. So, here is a trick I am sharing whenever you have to find the value of expression in which one variable is added or subtracted with a reciprocal of the same variable then equate it to some variable say t and then do the square on both the sides.
$\begin{align}
& x+\dfrac{1}{x}=t \\
& {{\left( x+\dfrac{1}{x} \right)}^{2}}={{t}^{2}} \\
& {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2.x.\dfrac{1}{x}={{t}^{2}} \\
\end{align}$
The advantage of doing this is that the coefficient of 2 becomes 1 so now it will be easier to solve.
Complete step-by-step answer:
Let us take $\sqrt{x}-\dfrac{1}{\sqrt{x}}$as t.
$\sqrt{x}-\dfrac{1}{\sqrt{x}}=t$
Squaring both the sides will change the above equation as follows:
${{\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)}^{2}}={{t}^{2}}$
Now, we are using${{\left( a-b \right)}^{2}}$identity in the above equation and we get,
$x+\dfrac{1}{x}-2={{t}^{2}}$
We are going to substitute the value of x which is equal to 3 + 2$\sqrt{2}$ in the above equation.$3+2\sqrt{2}+\dfrac{1}{3+2\sqrt{2}}-2={{t}^{2}}$
Rationalize the$\dfrac{1}{3+2\sqrt{2}}$. Rationalization is done by multiplying and dividing this rational number by the conjugate of 3 + 2$\sqrt{2}$ and conjugate of 3 + 2$\sqrt{2}$ is 3 -2$\sqrt{2}$. So, the rationalization will resolve the above expression as follows.
$3+2\sqrt{2}+\dfrac{3-2\sqrt{2}}{\left( 3+2\sqrt{2} \right)\left( 3-2\sqrt{2} \right)}-2={{t}^{2}}$
The expression in the denominator is $\left( a+b \right)\left( a-b \right)$ identity which is equal to ${{a}^{2}}-{{b}^{2}}$. So, using this identity in the above expression will give:
$\begin{align}
& 3+2\sqrt{2}+\dfrac{3-2\sqrt{2}}{{{\left( 3 \right)}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}}-2={{t}^{2}} \\
& \Rightarrow 3+2\sqrt{2}+\dfrac{3-2\sqrt{2}}{9-8}-2={{t}^{2}} \\
& \\
\end{align}$
$\Rightarrow 3+2\sqrt{2}+3-2\sqrt{2}-2={{t}^{2}}$
In the above equation, 2$\sqrt{2}$ and - 2$\sqrt{2}$ is cancelled out and the above expression will look as below:
$\begin{align}
& 6-2={{t}^{2}} \\
& \Rightarrow 4={{t}^{2}} \\
& \Rightarrow \pm 2=t \\
\end{align}$
From the above calculation, we have found the value of the expression $\sqrt{x}-\dfrac{1}{\sqrt{x}}$ as ±2.
Hence, the correct option is (b)
Note: If we just substitute the value of x in the expression $\sqrt{x}-\dfrac{1}{\sqrt{x}}$ it will look very complex to solve. So, here is a trick I am sharing whenever you have to find the value of expression in which one variable is added or subtracted with a reciprocal of the same variable then equate it to some variable say t and then do the square on both the sides.
$\begin{align}
& x+\dfrac{1}{x}=t \\
& {{\left( x+\dfrac{1}{x} \right)}^{2}}={{t}^{2}} \\
& {{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2.x.\dfrac{1}{x}={{t}^{2}} \\
\end{align}$
The advantage of doing this is that the coefficient of 2 becomes 1 so now it will be easier to solve.
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