Question

Rationalize the denominator and simplify:
$\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Answer 1

Hint:In the above given expression, rationalization of a denominator is done by multiplying and dividing the whole expression by the conjugate of $\sqrt{3}-\sqrt{2}$ which is equal to$\sqrt{3}+\sqrt{2}$. After rationalization, use the basic algebraic identities $(a + b)(a – b)=a^2 – b^2$ and ${(a+b)}^2=a^2+b^2+2ab$ simplify it and get the answer.

Complete step-by-step answer:
The expression given in the above question is:
$\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
To rationalize the denominator, we will multiply and divide the whole expression by $\sqrt{3}+\sqrt{2}$ which is the conjugate of $\sqrt{3}-\sqrt{2}$. Now, we are rationalizing the above expression as follows:
   $\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
In the above expression, the denominator is in the form of $(a + b)(a – b)$ which is equal to $a^2 – b^2$ and apply the identity ${(a+b)}^2=a^2+b^2+2ab$ in the numerator part, we get
Here $a={\sqrt3}$ and $b={\sqrt2}$
$\begin{align}
 & =\dfrac{{{\left( \sqrt{3}+\sqrt{2} \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \\
 & =\dfrac{{\sqrt3}^2+{\sqrt2}^2+2({\sqrt3})({\sqrt2})}{{\sqrt3}^2-{\sqrt2}^2} \\
 & =\dfrac{3+2+2\sqrt{6}}{3-2} \\
 & =\dfrac{5+2\sqrt{6}}{1} \\
\end{align}$
Hence, the rationalization followed by simplification of the given expression yields ${5+2\sqrt{6}}$

Note: You might be wondering as it is a question's requirement so we have rationalized this expression. But in general, why is there a need to rationalize? The answer is as you can see that after rationalization meaning multiplying and dividing the whole expression by a conjugate, the denominator of the expression is reduced to 1 by using basic algebraic identities.So, rationalization will simplify the denominator in such a way that it contains only rational numbers.So, in a calculation if you find the denominator can be rationalized then go for it, as it will reduce the complexity of the problem.