Question

Find the values of \[\dfrac{{{\left( \sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1} \right)}^{2}}\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}\]
1. \[\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}\]
2. \[1\]
3. \[\sqrt{3}\]
4. \[\dfrac{1}{\sqrt{3}}\]

Answer 1

Hint: According to the given problem first of all we have to solve the numerator by applying the property of \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] and simplifying the brackets in numerator and in denominator we can just simplify and get the answer. So, in this way we can solve the problem.

Complete step by step answer:
Given problem is that we have to find the value of\[\dfrac{{{\left( \sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1} \right)}^{2}}\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}--(1)\]
If you notice this problem we can solve by using the property of\[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] in numerator.
Here, in this problem we have \[a=\sqrt{\sqrt{3}+1}\,\,\] \[b=\sqrt{\sqrt{3}-1}\]and \[2ab=2\left( \sqrt{\sqrt{3}+1}\, \right)\left( \sqrt{\sqrt{3}-1} \right)\]
Now, we can apply this property in equation\[(1)\]
\[\dfrac{\left[ {{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}+2\left( \sqrt{\sqrt{3}+1} \right)\left( \sqrt{\sqrt{3}-1} \right)+{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}} \right]\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}\]
By modifying the numerator we get:
\[=\dfrac{\left[ \left( \sqrt{3}+1 \right)+2\left( \sqrt{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)} \right)+\left( \sqrt{3}-1 \right) \right]\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}\]
By using this property of \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]
Here in this problem\[a=\sqrt{\sqrt{3}+1}\,\,\] \[b=\sqrt{\sqrt{3}-1}\]
\[\left( \sqrt{\sqrt{3}+1} \right)\left( \sqrt{\sqrt{3}-1} \right)=\sqrt{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}=\sqrt{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}}}--(2)\]
Substitute the equation\[(2)\]in equation\[(1)\]
\[\dfrac{\left[ \left( \sqrt{3}+1 \right)+2\left( \sqrt{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}}} \right)+\left( \sqrt{3}-1 \right) \right]\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}\]
After simplifying further we get:
\[=\dfrac{\left[ 2\sqrt{3}+2\left( \sqrt{3-1} \right) \right]\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}\]
In the above equation take the 2 common in numerator
\[=\dfrac{2\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}\]
Now, in denominator take the squaring those brackets with roots and \[\sqrt{3}\] gets cancelled out remain 2
\[=\dfrac{2\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}{\sqrt{3}+1-\sqrt{3}+1}\]
After solving further we will get:
\[=\dfrac{2\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}{2}\]
Here in the equation 2 get cancelled and we get:
\[=\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)\]
If you kindly see this above equation in the form of \[(a-b)(a+b)\]
So we have to use the property of \[(a-b)(a+b)={{a}^{2}}-{{b}^{2}}\]
\[={{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}\]
After solving this squaring brackets
\[=3-2\]
After solving this you get:
\[=1\]
Hence, the value of \[\dfrac{{{\left( \sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1} \right)}^{2}}\left( \sqrt{3}-\sqrt{2} \right)}{{{\left( \sqrt{\sqrt{3}+1} \right)}^{2}}-{{\left( \sqrt{\sqrt{3}-1} \right)}^{2}}}=1\]

So, the correct answer is “Option 2”.

Note: In this particular problem students may be confused by seeing such complicated problems. But it’s actually easier. It can be advisable for such a problem to solve the brackets first by using the basic property of mathematics. Then the problems become easier to solve. So, in this way we can solve the problems.