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Question: Rationalize the denominator and simplify: \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)...

Rationalize the denominator and simplify:
3+23−2\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

Explanation

Solution

Hint:In the above given expression, rationalization of a denominator is done by multiplying and dividing the whole expression by the conjugate of 3−2\sqrt{3}-\sqrt{2} which is equal to3+2\sqrt{3}+\sqrt{2}. After rationalization, use the basic algebraic identities (a+b)(a–b)=a2–b2(a + b)(a – b)=a^2 – b^2 and (a+b)2=a2+b2+2ab{(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:
3+23−2\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}
To rationalize the denominator, we will multiply and divide the whole expression by 3+2\sqrt{3}+\sqrt{2} which is the conjugate of 3−2\sqrt{3}-\sqrt{2}. Now, we are rationalizing the above expression as follows:
3+23−2×3+23+2\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)(a + b)(a – b) which is equal to a2–b2a^2 – b^2 and apply the identity (a+b)2=a2+b2+2ab{(a+b)}^2=a^2+b^2+2ab in the numerator part, we get
Here a=3a={\sqrt3} and b=2b={\sqrt2}
=(3+2)2(3)2−(2)2 =32+22+2(3)(2)32−22 =3+2+263−2 =5+261 \begin{aligned} & =\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{aligned}
Hence, the rationalization followed by simplification of the given expression yields 5+26{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.