Question : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Option 1: $1$
Option 2: $2$
Option 3: $\sqrt{3}$
Option 4: $\sqrt{5}$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $2$
Solution :
Given: $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
Now, $\frac{x}{\sqrt{20}}=\frac{4\sqrt{5}\sqrt{3}}{\sqrt{20}(\sqrt{5}+\sqrt{3})} = \frac{2\sqrt{3}}{(\sqrt{5}+\sqrt{3})}$
Using componendo and dividendo
$\frac{x+\sqrt{20}}{x-\sqrt{20}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}$
Similarly, we can find $\frac{x+\sqrt{12}}{x-\sqrt{12}}=\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})}$
Now, $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}=\frac{\sqrt{5}+3\sqrt{3}}{(\sqrt{3}-\sqrt{5})}+\frac{\sqrt{3}+3\sqrt{5}}{(\sqrt{5}-\sqrt{3})} =\frac{2\sqrt{3}-2\sqrt{5}}{(\sqrt{3}-\sqrt{5})}$ =2
Hence, the correct answer is $2$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
