Question

If $ x=3+2\sqrt{2} $ , then find the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ ?
(a) $ \pm 2 $
(b) $ \pm 4 $
(c) $ \pm 3 $
(d) none of the above

Answer 1

Hint: We start solving the problem by finding the value of $ \sqrt{x} $ by converting the given value of x resembling $ {{a}^{2}}+{{b}^{2}}+2ab $ . We then simplify the given $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ and substitute the obtained value of $ \sqrt{x} $ and x in it. We then make the necessary calculations to get the required value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ .

Complete step by step answer:
According to the problem, we are given that $ x=3+2\sqrt{2} $ and we need to find the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ .
Let us first find the value of $ \sqrt{x} $ .
We have $ x=3+2\sqrt{2} $ .
 $ \Rightarrow x=2+1+2\sqrt{2} $ .
 $ \Rightarrow x={{\left( \sqrt{2} \right)}^{2}}+{{\left( 1 \right)}^{2}}+2\left( \sqrt{2} \right)\left( 1 \right) $ ---(1).
We can see that equation (1) resembles the equation $ {{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}} $ .
So, we get $ x={{\left( \sqrt{2}+1 \right)}^{2}} $ .
 $ \Rightarrow \sqrt{x}=\pm \left( \sqrt{2}+1 \right) $ .
Let us assume we use these values to find the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ .
So, we get $ \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{{{\left( \sqrt{x} \right)}^{2}}-1}{\sqrt{x}} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}} $ .
Now, let us substitute the values of $ \sqrt{x} $ and x.
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{3+2\sqrt{2}-1}{\pm \left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{2\sqrt{2}+2}{\pm \left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{2\left( \sqrt{2}+1 \right)}{\pm \left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\pm 2 $ .
So, we have found the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ as $ \pm 2 $ .

$ \therefore $ The correct option for the given problem is (a).

Note:
 Whenever we get this type of problem, we should first convert the given value of x into a square. We can also solve the problem after finding the value of $ \sqrt{x} $ as shown below:
Let us assume $ \sqrt{x}=+\left( \sqrt{2}+1 \right) $ to find the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ .
So, we have $ \sqrt{x}-\dfrac{1}{\sqrt{x}}=\left( \sqrt{2}+1 \right)-\dfrac{1}{\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{{{\left( \sqrt{2}+1 \right)}^{2}}-1}{\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{{{\left( \sqrt{2} \right)}^{2}}+{{\left( 1 \right)}^{2}}+2\left( \sqrt{2} \right)\left( 1 \right)-1}{\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{2+1+2\sqrt{2}-1}{\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{2+2\sqrt{2}}{\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=2 $ .
Let us assume $ \sqrt{x}=-\left( \sqrt{2}+1 \right) $ to find the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ .
So, we have $ \sqrt{x}-\dfrac{1}{\sqrt{x}}=-\left( \sqrt{2}+1 \right)-\left( \dfrac{1}{-\left( \sqrt{2}+1 \right)} \right) $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{{{\left( \sqrt{2}+1 \right)}^{2}}-1}{-\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{{{\left( \sqrt{2} \right)}^{2}}+{{\left( 1 \right)}^{2}}+2\left( \sqrt{2} \right)\left( 1 \right)-1}{-\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{2+1+2\sqrt{2}-1}{-\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{2+2\sqrt{2}}{-\left( \sqrt{2}+1 \right)} $ .
 $ \Rightarrow \sqrt{x}-\dfrac{1}{\sqrt{x}}=-2 $ .
So, the value of $ \sqrt{x}-\dfrac{1}{\sqrt{x}} $ is $ \pm 2 $ .