Simplify: frac{2}{sqrt{5}-sqrt{3}} + frac{4}{ | vedclass

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

Question

(B) Step $1$: Rationalize the first term $\frac{2}{\sqrt{5}-\sqrt{3}}$.Multiply numerator and denominator by $(\sqrt{5} + \sqrt{3})$:$\frac{2(\sqrt{5}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Step $1$: Rationalize the first term $\frac{2}{\sqrt{5}-\sqrt{3}}$.Multiply numerator and denominator by $(\sqrt{5} + \sqrt{3})$:$\frac{2(\sqrt{5} + \sqrt{3})}{5 - 3} = \frac{2(\sqrt{5} + \sqrt{3})}{2} = \sqrt{5} + \sqrt{3}$.Step $2$: Simplify the second term $\frac{4}{\sqrt{10+\sqrt{84}}}$.Note that $\sqrt{84} = \sqrt{4 	imes 21} = 2\sqrt{21}$.So,$\sqrt{10 + 2\sqrt{21}} = \sqrt{(\sqrt{7} + \sqrt{3})^2} = \sqrt{7} + \sqrt{3}$.Thus,the term becomes $\frac{4}{\sqrt{7} + \sqrt{3}}$.Rationalize it: $\frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$.Step $3$: Rationalize the third term $\frac{2}{\sqrt{7}-\sqrt{5}}$.Multiply numerator and denominator by $(\sqrt{7} + \sqrt{5})$:$\frac{2(\sqrt{7} + \sqrt{5})}{7 - 5} = \frac{2(\sqrt{7} + \sqrt{5})}{2} = \sqrt{7} + \sqrt{5}$.Step $4$: Combine all terms:$(\sqrt{5} + \sqrt{3}) + (\sqrt{7} - \sqrt{3}) - (\sqrt{7} + \sqrt{5})$$= \sqrt{5} + \sqrt{3} + \sqrt{7} - \sqrt{3} - \sqrt{7} - \sqrt{5} = 0$.

Simplify: $\frac{2}{\sqrt{5}-\sqrt{3}} + \frac{4}{\sqrt{10+\sqrt{84}}} - \frac{2}{\sqrt{7}-\sqrt{5}}$

Similar Questions

Can two numbers have $18$ as their $HCF$ and $380$ as their $LCM$? Give reasons.

Prove that the square of any odd positive integer is of the form $8m + 1$,where $m$ is a non-negative integer.

Explain why $3 \times 5 \times 7 + 7$ is a composite number.

If $k_{1}$ and $k_{2}$ are two distinct prime integers,then their $\text{l.c.m.}$ is ..........

For some integer $m$,every even integer is of the form

Vedclass Test Series

Exam Paper Generator

Online Exam Module