If frac{1}{N} = frac{(sqrt{6} + sqrt{5})}{(sq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\frac{1}{N} = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}.$To find $N,$ we take the reciprocal of both sides: $N = \frac{\sqrt{6}

Vedclass · Accepted Answer

$11 - 2\sqrt{30}$

Vedclass · Answer

(C) Given that $\frac{1}{N} = \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}}.$To find $N,$ we take the reciprocal of both sides: $N = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}}.$Now,rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{6} - \sqrt{5})$:$N = \frac{(\sqrt{6} - \sqrt{5})(\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})}$Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$ for the numerator and $(a + b)(a - b) = a^2 - b^2$ for the denominator:$N = \frac{(\sqrt{6})^2 + (\sqrt{5})^2 - 2(\sqrt{6})(\sqrt{5})}{(\sqrt{6})^2 - (\sqrt{5})^2}$$N = \frac{6 + 5 - 2\sqrt{30}}{6 - 5}$$N = \frac{11 - 2\sqrt{30}}{1}$$N = 11 - 2\sqrt{30}.$

If $\frac{1}{N} = \frac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} - \sqrt{5})},$ what is the value of $N?$

Similar Questions

Between two fractions $\frac{1}{2}$ and $\frac{1}{8},$ how many fractions are there in all?

$\frac{1}{3} + \frac{1}{2} + \frac{1}{x} = 4$. Then $x = ?$

If $2x + \frac{2}{x} = 1$,then the value of $x^{3} + \frac{1}{x^{3}}$ is

$(0 . \overline{09} \times 7 . \overline{3})$ is equal to

If $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}=4,$ then the value of $a^{2}+b^{2}$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module