Question

Solve and simplify the following expression,
$\dfrac{13}{\sqrt{27-2\sqrt{140}}}-\dfrac{2}{\sqrt{12-2\sqrt{35}}}$
(a) $\sqrt{5}$
(b) 2$\sqrt{5}$
(c) 3$\sqrt{5}$
(d) 4$\sqrt{5}$

Answer 1

Hint: To solve this problem we will first try to convert the denominator of both the given fractions into a perfect square so that we can cancel out the square with the root and get a simple equation. After that we will rationalize both the obtained fractions and further solve them and simply them according to the options given to get the required answer.

Complete step-by-step answer:
To solve this first we will observe the given expression i.e.
$\dfrac{13}{\sqrt{27-2\sqrt{140}}}-\dfrac{2}{\sqrt{12-2\sqrt{35}}}$
Now we can see that in the denominator part of both the above fractions there are complex equations involving square roots so to simplify it we will try to make the perfect square of the denominators of both the fractions so that we can cancel out the square with the root and leave with a simpler equation. This can be done as,
First we will convert $\sqrt{27-2\sqrt{140}}$ to perfect square,
So we get,
$\sqrt{27-2\sqrt{140}}$
This can also be written as,
$=\sqrt{{{\left( \sqrt{20} \right)}^{2}}+{{\left( \sqrt{7} \right)}^{2}}-2\sqrt{20}.\sqrt{7}}$
Now using the identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ in the above expression we get,
$=\sqrt{{{\left( \sqrt{20}-\sqrt{7} \right)}^{2}}}$
Cancelling out square with root we get,
$=\sqrt{20}-\sqrt{7}$
Now coming to the second denominator we have $\sqrt{12-2\sqrt{35}}$, so we will solve this as,
$\sqrt{12-2\sqrt{35}}$
This can also be written as,
$=\sqrt{{{\left( \sqrt{7} \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}-2\sqrt{7}.\sqrt{5}}$
Now using the identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ in the above expression we get,
$=\sqrt{{{\left( \sqrt{7}-\sqrt{5} \right)}^{2}}}$
Cancelling out square with root we get,
$=\sqrt{7}-\sqrt{5}$
So now we can represent the given equation as,
$\dfrac{13}{\sqrt{20}-\sqrt{7}}-\dfrac{2}{\sqrt{7}-\sqrt{5}}$
Now rationalizing the above equation’s fractions we get,
$=\dfrac{13}{\sqrt{20}-\sqrt{7}}\times \dfrac{\sqrt{20}+\sqrt{7}}{\sqrt{20}+\sqrt{7}}-\dfrac{2}{\sqrt{7}-\sqrt{5}}\times \dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, using this identity in above expression we get,
\[\begin{align}
  & =\dfrac{13\left( \sqrt{20}+\sqrt{7} \right)}{{{\left( \sqrt{20} \right)}^{2}}-{{\left( \sqrt{7} \right)}^{2}}}-\dfrac{2\left( \sqrt{7}+\sqrt{5} \right)}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}} \\
 & =\dfrac{13\left( \sqrt{20}+\sqrt{7} \right)}{20-7}-\dfrac{2\left( \sqrt{7}+\sqrt{5} \right)}{7-5} \\
 & =\dfrac{13\left( \sqrt{20}+\sqrt{7} \right)}{13}-\dfrac{2\left( \sqrt{7}+\sqrt{5} \right)}{2} \\
 & =\left( \sqrt{20}+\sqrt{7} \right)-\left( \sqrt{7}+\sqrt{5} \right) \\
 & =\sqrt{20}+\sqrt{7}-\sqrt{7}-\sqrt{5} \\
 & =\sqrt{20}-\sqrt{5} \\
\end{align}\]
We can also write $\sqrt{20}$ as $2\sqrt{5}$ so,
$\begin{align}
  & =2\sqrt{5}-\sqrt{5} \\
 & =\sqrt{5} \\
\end{align}$

So, the correct answer is “Option A”.

Note: Whenever we are given complex equations in the denominator part of the fraction which involves square roots try to simplify it first by either converting it to perfect square or by rationalizing the fraction. And also be careful while making the calculation as there are many chances of committing calculation mistakes in these types of questions.