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: $13$
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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