Question

If \[x=7-4\sqrt{3}\], then find the value of
(i) \[\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\]
(ii) \[\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)\]

Answer 1

Hint: First of all express, \[x=7-4\sqrt{3}\text{ as }x={{\left( \sqrt{3} \right)}^{2}}+{{\left( 2 \right)}^{2}}-2.2\sqrt{3}\] which can be converted into a perfect square by using \[{{a}^{2}}+{{b}^{2}}-2ab={{\left( a+b \right)}^{2}}\]. From here, find the value of \[\sqrt{x}\text{ and }\dfrac{1}{\sqrt{x}}\] and substitute these values in \[\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\text{ and }\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)\] to find the correct answer.

Complete step by step solution:
We are given that \[x=7-4\sqrt{3}\], then we have to find the value of
(i) \[\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\]
(ii) \[\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)\]
Let us consider the value of x given in the question
\[x=7-4\sqrt{3}\]
We can write the above value of x as
\[x=3+4-4\sqrt{3}\]
\[x={{\left( \sqrt{3} \right)}^{2}}+{{\left( 2 \right)}^{2}}-2.2\sqrt{3}\]
We know that,
\[{{a}^{2}}+{{b}^{2}}-2ab={{\left( a+b \right)}^{2}}\]
So, by using this and considering \[a=\sqrt{3}\] and b = 2, we get
\[x={{\left( \sqrt{3}-2 \right)}^{2}}\]
By taking square root on both the sides of the above equation, we get,
\[\sqrt{x}=\sqrt{{{\left( \sqrt{3}-2 \right)}^{2}}}\]
So, we get,
\[\sqrt{x}=\pm \left( \sqrt{3}-2 \right)\]
\[\sqrt{x}=\left( \sqrt{3}-2 \right);\sqrt{x}=-\left( \sqrt{3}-2 \right)\]
Also, for \[\sqrt{x}=\left( \sqrt{3}-2 \right)\], we get \[\dfrac{1}{\sqrt{x}}=\dfrac{1}{\left( \sqrt{3}-2 \right)}\]
By multiplying \[\left( \sqrt{3}+2 \right)\] on both the numerator and denominator, we get,
\[\dfrac{1}{\sqrt{x}}=\dfrac{\left( \sqrt{3}+2 \right)}{\left( \sqrt{3}-2 \right)\left( \sqrt{3}+2 \right)}\]
By using \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\], we get,
\[\dfrac{1}{\sqrt{x}}=\dfrac{\left( \sqrt{3}+2 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 2 \right)}^{2}}}\]
\[\dfrac{1}{\sqrt{x}}=\dfrac{\left( \sqrt{3}+2 \right)}{3-4}\]
\[\dfrac{1}{\sqrt{x}}=-\left( \sqrt{3}+2 \right)\]
So, we get,
\[\sqrt{x}=\left( \sqrt{3}-2 \right)\text{ and }\dfrac{1}{\sqrt{x}}=-\left( \sqrt{3}+2 \right)....\left( i \right)\]
Now, for \[\sqrt{x}=-\left( \sqrt{3}-2 \right)\], we get,
\[\dfrac{1}{\sqrt{x}}=\dfrac{1}{-\left( \sqrt{3}-2 \right)}\]
By multiplying, \[\left( \sqrt{3}+2 \right)\] on both numerator and denominator, we get,
\[\dfrac{1}{\sqrt{x}}=\dfrac{\left( \sqrt{3}+2 \right)}{-\left( \sqrt{3}-2 \right)\left( \sqrt{3}+2 \right)}\]
\[\dfrac{1}{\sqrt{x}}=\dfrac{\left( \sqrt{3}+2 \right)}{-\left[ {{\left( \sqrt{3} \right)}^{2}}-{{\left( 2 \right)}^{2}} \right]}\]
\[\dfrac{1}{\sqrt{x}}=\dfrac{\left( \sqrt{3}+2 \right)}{-\left( 3-4 \right)}\]
\[\dfrac{1}{\sqrt{x}}=\left( \sqrt{3}+2 \right)\]
So, we get,
\[\sqrt{x}=-\left( \sqrt{3}-2 \right)\text{ and }\dfrac{1}{\sqrt{x}}=\left( \sqrt{3}+2 \right)....\left( ii \right)\]
Now, let us find the value of
(i) \[E=\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\]
From equation (i), we have
\[\sqrt{x}=\left( \sqrt{3}-2 \right)\text{ and }\dfrac{1}{\sqrt{x}}=-\left( \sqrt{3}+2 \right)\]
So, substituting these in the above expression, we get,
\[E=\left( \sqrt{3}-2 \right)+\left[ -\left( \sqrt{3}+2 \right) \right]\]
\[E=\sqrt{3}-2-\sqrt{3}-2\]
\[E=-4\]
From equation (ii), we have
\[\sqrt{x}=-\left( \sqrt{3}-2 \right)\text{ and }\dfrac{1}{\sqrt{x}}=\left( \sqrt{3}+2 \right)\]
So, substituting these in the above expression, we get,
\[E=\left( \sqrt{3}-2 \right)+\left( \sqrt{3}+2 \right)\]
\[E=-\sqrt{3}+2+\sqrt{3}-2\]
\[E=4\]
Now, let us find the value of
(ii) \[F=\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)\]
From equation (i), we have
\[\sqrt{x}=\left( \sqrt{3}-2 \right)\text{ and }\dfrac{1}{\sqrt{x}}=-\left( \sqrt{3}+2 \right)\]
So, substituting these in the above expression, we get,
\[F=\left( \sqrt{3}-2 \right)-\left[ -\left( \sqrt{3}+2 \right) \right]\]
\[F=\sqrt{3}-2+\sqrt{3}+2\]
\[F=2\sqrt{3}\]
From equation (ii), we have
\[\sqrt{x}=-\left( \sqrt{3}-2 \right)\text{ and }\dfrac{1}{\sqrt{x}}=\left( \sqrt{3}+2 \right)\]
So, substituting these in the above expression, we get,
\[F=-\left( \sqrt{3}-2 \right)-\left( \sqrt{3}+2 \right)\]
\[F=-\sqrt{3}+2-\sqrt{3}-2\]
\[F=-2\sqrt{3}\]

Note: In this question, many students just take \[\sqrt{x}=\sqrt{3}-2\] and they miss \[\sqrt{x}=-\left( \sqrt{3}-2 \right)\], but both the values of \[\sqrt{x}\] should be taken. So, this must be taken care of. Also, students must take \[\dfrac{1}{\sqrt{x}}\] corresponding to the given \[\sqrt{x}\] and should not get confused between the two values of \[\sqrt{x}\] by doing it separately. Also, students should always rationalize the denominator whenever it is irrational by multiplying it with its conjugate like for example, we convert \[\dfrac{1}{\left( \sqrt{3}-2 \right)}\text{ to }-\left( \sqrt{3}+2 \right)\] by multiplying both numerator and denominator by \[\left( \sqrt{3}+2 \right)\].