If $x=2+\sqrt{3}$ and $xy=1$, then $\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}=?$
(a) $\sqrt{2}$
(b) $\sqrt{3}$
(c) 1
(d) None of these
Answer
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Hint: Find the value of y using the two relations between x and y. Rationalize the term $2+\sqrt{3}$ by multiplying the numerator and the denominator with its conjugate $2-\sqrt{3}$ to get the value of y. Now, assume the given expression whose value we have to find as E and rationalize its both the terms. Use the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to simplify. Further, write the expression \[\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right)={{\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right)}^{3}}-3{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right)\] using the formula $\left( {{a}^{3}}+{{b}^{3}} \right)={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$. To find the value of ${{x}^{\dfrac{1}{2}}}$, multiply $x=2+\sqrt{3}$ with 2 both the sides and write $\left( 4+\sqrt{3} \right)={{\left( \sqrt{3}+1 \right)}^{2}}$ and take square root both the sides. Similarly, find the value of ${{y}^{\dfrac{1}{2}}}$ and substitute in the expression to get the answer.
Complete step by step answer:
Here we have been provided with the relations $x=2+\sqrt{3}$ and $xy=1$, we are asked to find the value of the expression $\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}$. First we need to find the value of y. Let us assume the expression as E, so we have,
$\Rightarrow E=\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}$
Now, using the relation $xy=1$ we can write $y=\dfrac{1}{x}$, so substituting the value of x we get,
$\Rightarrow y=\dfrac{1}{2+\sqrt{3}}$
Rationalizing the denominator of the above term by multiplying both the numerator and the denominator with the conjugate of $2+\sqrt{3}$ which is $2-\sqrt{3}$ we get,
$\Rightarrow y=\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}$
Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we get,
$\begin{align}
& \Rightarrow y=\dfrac{\left( 2-\sqrt{3} \right)}{\left( {{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)} \\
& \Rightarrow y=\dfrac{\left( 2-\sqrt{3} \right)}{4-3} \\
& \Rightarrow y=2-\sqrt{3} \\
\end{align}$
Now let us simplify the expression E using by rationalizing its both the terms, so the numerator and the denominator of the first term should be multiplied with $\sqrt{2}-\sqrt{x}$ and similarly the numerator and the denominator of the second term should be multiplied with $\sqrt{2}+\sqrt{y}$. So we get,
$\begin{align}
& \Rightarrow E=\dfrac{x}{\sqrt{2}+\sqrt{x}}\times \dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{2}-\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}\times \dfrac{\sqrt{2}+\sqrt{y}}{\sqrt{2}+\sqrt{y}} \\
& \Rightarrow E=\dfrac{\sqrt{2}x-x\sqrt{x}}{2-x}+\dfrac{\sqrt{2}y+y\sqrt{y}}{2-y} \\
\end{align}$
Using the relations $x=2+\sqrt{3}$ and $y=2-\sqrt{3}$ we get,
$\begin{align}
& \Rightarrow E=\dfrac{\sqrt{2}x-{{x}^{\dfrac{3}{2}}}}{-\sqrt{3}}+\dfrac{\sqrt{2}y+{{y}^{\dfrac{3}{2}}}}{\sqrt{3}} \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -\sqrt{2}x+{{x}^{\dfrac{3}{2}}}+\sqrt{2}y+{{y}^{\dfrac{3}{2}}} \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( \sqrt{2}\left( y-x \right)+\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right) \right) \\
\end{align}$
Substituting the values of x and y we get,
$\begin{align}
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( \sqrt{2}\left( 2-\sqrt{3}-2-\sqrt{3} \right)+\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right) \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right) \right) \\
\end{align}$
Using the algebraic identity $\left( {{a}^{3}}+{{b}^{3}} \right)={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ we can simplify the above expression as,
$\Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right)}^{3}}-3{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right) \right)$
We have $xy=1$ so taking square root both the sides we get ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}=1$, therefore we get,
$\Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right)}^{3}}-3\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right) \right)$ ……. (1)
We need to find the values of ${{x}^{\dfrac{1}{2}}}$ and ${{y}^{\dfrac{1}{2}}}$. So we have,
$\Rightarrow x=2+\sqrt{3}$
Multiplying both the sides with 2 we get,
$\begin{align}
& \Rightarrow 2x=4+2\sqrt{3} \\
& \Rightarrow 2x={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}+2\sqrt{3} \\
\end{align}$
The above relation is of the form ${{a}^{2}}+{{b}^{2}}+2ab$ whose simplified form is given as ${{\left( a+b \right)}^{2}}$, so we get,
$\begin{align}
& \Rightarrow 2x={{\left( \sqrt{3}+1 \right)}^{2}} \\
& \Rightarrow x=\dfrac{{{\left( \sqrt{3}+1 \right)}^{2}}}{2} \\
\end{align}$
Taking square root both the sides we get,
$\Rightarrow {{x}^{\dfrac{1}{2}}}=\dfrac{\left( \sqrt{3}+1 \right)}{\sqrt{2}}$
Similarly, we can find the value of ${{y}^{\dfrac{1}{2}}}$ using the similar process, so we get,
$\begin{align}
& \Rightarrow 2y=4-2\sqrt{3} \\
& \Rightarrow 2y={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}-2\sqrt{3} \\
\end{align}$
The above relation is of the form ${{a}^{2}}+{{b}^{2}}-2ab$ whose simplified form is given as ${{\left( a-b \right)}^{2}}$, so we get,
$\begin{align}
& \Rightarrow 2y={{\left( \sqrt{3}-1 \right)}^{2}} \\
& \Rightarrow y=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{2} \\
\end{align}$
Taking the square root both the sides we get,
$\Rightarrow {{y}^{\dfrac{1}{2}}}=\dfrac{\left( \sqrt{3}-1 \right)}{\sqrt{2}}$
Now, substituting the values of ${{x}^{\dfrac{1}{2}}}$ and ${{y}^{\dfrac{1}{2}}}$ in expression (1) we get,
$\begin{align}
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( \left( \dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}} \right) \right)}^{3}}-3\left( \left( \dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}} \right) \right) \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( \sqrt{6} \right)}^{3}}-3\left( \sqrt{6} \right) \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( 6\sqrt{6}-5\sqrt{6} \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\times \left( \sqrt{6} \right) \\
\end{align}$
We can write $\sqrt{6}=\sqrt{2}\times \sqrt{3}$ so cancelling the common factors we get,
$\therefore E=\sqrt{2}$
So, the correct answer is “Option a”.
Note: Note that it is important to rationalize the given terms otherwise the calculations will become difficult. Do not substitute the decimal value of $\sqrt{3}$ to simplify as it will make the expression complicated to calculate while using the algebraic identities. Remember all the algebraic identities to make the simplification easier. In the rationalization process both the numerator and denominator are multiplied with the conjugate of the denominator or numerator as per the requirement.
Complete step by step answer:
Here we have been provided with the relations $x=2+\sqrt{3}$ and $xy=1$, we are asked to find the value of the expression $\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}$. First we need to find the value of y. Let us assume the expression as E, so we have,
$\Rightarrow E=\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}$
Now, using the relation $xy=1$ we can write $y=\dfrac{1}{x}$, so substituting the value of x we get,
$\Rightarrow y=\dfrac{1}{2+\sqrt{3}}$
Rationalizing the denominator of the above term by multiplying both the numerator and the denominator with the conjugate of $2+\sqrt{3}$ which is $2-\sqrt{3}$ we get,
$\Rightarrow y=\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}$
Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we get,
$\begin{align}
& \Rightarrow y=\dfrac{\left( 2-\sqrt{3} \right)}{\left( {{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)} \\
& \Rightarrow y=\dfrac{\left( 2-\sqrt{3} \right)}{4-3} \\
& \Rightarrow y=2-\sqrt{3} \\
\end{align}$
Now let us simplify the expression E using by rationalizing its both the terms, so the numerator and the denominator of the first term should be multiplied with $\sqrt{2}-\sqrt{x}$ and similarly the numerator and the denominator of the second term should be multiplied with $\sqrt{2}+\sqrt{y}$. So we get,
$\begin{align}
& \Rightarrow E=\dfrac{x}{\sqrt{2}+\sqrt{x}}\times \dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{2}-\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}\times \dfrac{\sqrt{2}+\sqrt{y}}{\sqrt{2}+\sqrt{y}} \\
& \Rightarrow E=\dfrac{\sqrt{2}x-x\sqrt{x}}{2-x}+\dfrac{\sqrt{2}y+y\sqrt{y}}{2-y} \\
\end{align}$
Using the relations $x=2+\sqrt{3}$ and $y=2-\sqrt{3}$ we get,
$\begin{align}
& \Rightarrow E=\dfrac{\sqrt{2}x-{{x}^{\dfrac{3}{2}}}}{-\sqrt{3}}+\dfrac{\sqrt{2}y+{{y}^{\dfrac{3}{2}}}}{\sqrt{3}} \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -\sqrt{2}x+{{x}^{\dfrac{3}{2}}}+\sqrt{2}y+{{y}^{\dfrac{3}{2}}} \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( \sqrt{2}\left( y-x \right)+\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right) \right) \\
\end{align}$
Substituting the values of x and y we get,
$\begin{align}
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( \sqrt{2}\left( 2-\sqrt{3}-2-\sqrt{3} \right)+\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right) \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+\left( {{x}^{\dfrac{3}{2}}}+{{y}^{\dfrac{3}{2}}} \right) \right) \\
\end{align}$
Using the algebraic identity $\left( {{a}^{3}}+{{b}^{3}} \right)={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ we can simplify the above expression as,
$\Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right)}^{3}}-3{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right) \right)$
We have $xy=1$ so taking square root both the sides we get ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}=1$, therefore we get,
$\Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right)}^{3}}-3\left( {{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}} \right) \right)$ ……. (1)
We need to find the values of ${{x}^{\dfrac{1}{2}}}$ and ${{y}^{\dfrac{1}{2}}}$. So we have,
$\Rightarrow x=2+\sqrt{3}$
Multiplying both the sides with 2 we get,
$\begin{align}
& \Rightarrow 2x=4+2\sqrt{3} \\
& \Rightarrow 2x={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}+2\sqrt{3} \\
\end{align}$
The above relation is of the form ${{a}^{2}}+{{b}^{2}}+2ab$ whose simplified form is given as ${{\left( a+b \right)}^{2}}$, so we get,
$\begin{align}
& \Rightarrow 2x={{\left( \sqrt{3}+1 \right)}^{2}} \\
& \Rightarrow x=\dfrac{{{\left( \sqrt{3}+1 \right)}^{2}}}{2} \\
\end{align}$
Taking square root both the sides we get,
$\Rightarrow {{x}^{\dfrac{1}{2}}}=\dfrac{\left( \sqrt{3}+1 \right)}{\sqrt{2}}$
Similarly, we can find the value of ${{y}^{\dfrac{1}{2}}}$ using the similar process, so we get,
$\begin{align}
& \Rightarrow 2y=4-2\sqrt{3} \\
& \Rightarrow 2y={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}-2\sqrt{3} \\
\end{align}$
The above relation is of the form ${{a}^{2}}+{{b}^{2}}-2ab$ whose simplified form is given as ${{\left( a-b \right)}^{2}}$, so we get,
$\begin{align}
& \Rightarrow 2y={{\left( \sqrt{3}-1 \right)}^{2}} \\
& \Rightarrow y=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{2} \\
\end{align}$
Taking the square root both the sides we get,
$\Rightarrow {{y}^{\dfrac{1}{2}}}=\dfrac{\left( \sqrt{3}-1 \right)}{\sqrt{2}}$
Now, substituting the values of ${{x}^{\dfrac{1}{2}}}$ and ${{y}^{\dfrac{1}{2}}}$ in expression (1) we get,
$\begin{align}
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( \left( \dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}} \right) \right)}^{3}}-3\left( \left( \dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}} \right) \right) \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( -2\sqrt{6}+{{\left( \sqrt{6} \right)}^{3}}-3\left( \sqrt{6} \right) \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\left( 6\sqrt{6}-5\sqrt{6} \right) \\
& \Rightarrow E=\dfrac{1}{\sqrt{3}}\times \left( \sqrt{6} \right) \\
\end{align}$
We can write $\sqrt{6}=\sqrt{2}\times \sqrt{3}$ so cancelling the common factors we get,
$\therefore E=\sqrt{2}$
So, the correct answer is “Option a”.
Note: Note that it is important to rationalize the given terms otherwise the calculations will become difficult. Do not substitute the decimal value of $\sqrt{3}$ to simplify as it will make the expression complicated to calculate while using the algebraic identities. Remember all the algebraic identities to make the simplification easier. In the rationalization process both the numerator and denominator are multiplied with the conjugate of the denominator or numerator as per the requirement.
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