pharm wrote:
I have a quick question on this ..when the initial fraction was rationalized you used:
\(\sqrt{n+1}+ \sqrt{n} / \sqrt{n+1}+ \sqrt{n}\)
did you change the sign from negative to positive since the question stated "n" is a positive number. Wouldn't you have to use the same denominator when Rationalizing a fraction?
When there is an irrational number in the denominator, you rationalize it by multiplying it with its complement i.e. if it is \(\sqrt{a} + \sqrt{b}\) in the denominator, you will multiply by \(\sqrt{a} - \sqrt{b}\). This is done to use the algebraic identity (a + b)(a - b) = a^2 - b^2. When a and b are irrational, a^2 and b^2 become rational (given we are dealing with only square roots)
To keep the fraction same, you need to multiply the numerator with the same number as well.
An example will make it clear:
Rationalize
\(\frac{3}{{\sqrt{2} - 1}}\)
= \(\frac{3}{{\sqrt{2} - 1}} * \frac{\sqrt{2} + 1}{\sqrt{2} + 1}\)
= \(\frac{3*(\sqrt{2} + 1)}{(\sqrt{2})^2 - 1^2}\)
= \(\frac{3*(\sqrt{2} + 1)}{2 - 1}\)
The denominator has become rational.
Similarly, if the denominator has \(\sqrt{a} - \sqrt{b}\), you will multiply by \(\sqrt{a} + \sqrt{b}\).
In this question too, you can substitute n = 1. The given expression becomes \(\frac{1}{{\sqrt{2} - 1}}\)
Rationalize it and you will get \(\sqrt{2} + 1\). Put n = 1 in the options. Only option (E) gives you \(\sqrt{2} + 1\).
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