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Question : If $N=\frac{(\sqrt7-\sqrt3)}{(\sqrt7+\sqrt3)}$, then what is the value of $(N+\frac{1}{N})$?

Option 1: $2\sqrt2$

Option 2: $5$

Option 3: $10$

Option 4: $13$


Team Careers360 10th Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

Correct Answer: $5$


Solution : $N=\frac{(\sqrt7-\sqrt3)}{(\sqrt7+\sqrt3)}$, then what is the value of $N+(\frac{1}{N})$
$\frac{1}{N}=\frac{(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt3)}$
To find: $N+(\frac{1}{N})$
Putting the values, we get:
$\frac{(\sqrt7-\sqrt3)}{(\sqrt7+\sqrt3)}+\frac{(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt3)}$
= $\frac{(\sqrt7-\sqrt3)^2+(\sqrt7+\sqrt3)^2}{(\sqrt7+\sqrt3)(\sqrt7-\sqrt3)}$
= $\frac{(7+3-2\sqrt7\sqrt3)+(7+3+2\sqrt7\sqrt3)}{(7-3)}$
= $\frac{20}{4}$
= 5
Hence, the correct answer is $5$.

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