Question : If $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$, what is the value of $x$?
Option 1: $\frac{5}{2}$
Option 2: $\frac{25}{3}$
Option 3: $4$
Option 4: $3$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $3$
Solution : Given: $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$ Take $a=(\sqrt{5+x}+\sqrt{5-x})$, $b=(\sqrt{5+x}-\sqrt{5-x}), m=3$ and $n=1$. Applying the rule of componendo and dividendo, $\frac{a}{b}=\frac{m}{n}$ ⇒ $\frac{a+b}{a-b}=\frac{m+n}{m-n}$ Now, $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$ ⇒ $\frac{\sqrt{5+x}+\sqrt{5-x}+\sqrt{5+x}-\sqrt{5-x}}{\sqrt{5+x}+\sqrt{5-x}- \sqrt{5+x}+\sqrt{5-x}}=\frac{3+1}{3-1}$ ⇒ $\frac{\sqrt{5+x}+\sqrt{5+x}}{\sqrt{5-x}+\sqrt{5-x}}=\frac{4}{2}$ ⇒ $\frac{2\sqrt{5+x}}{2\sqrt{5-x}}=\frac{4}{2}$ ⇒ $\frac{\sqrt{5+x}}{\sqrt{5-x}}=2$ Squaring both sides, we get, ⇒ $\left(\frac{\sqrt{5+x}}{\sqrt{5-x}}\right)^{2}=2^{2}$ ⇒ $\frac{5+x}{5-x}=4$ ⇒ $5+x=20-4x$ ⇒ $5x=15$ $\therefore x=3$ Hence, the correct answer is $3$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
Question : If $x-\frac{1}{x}=1$, then what is the value of $\left (\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^{2}+1}-\frac{1}{x^{2}-1} \right)\;$?
Question : If $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$
Question : If $x=\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$ and $y=\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}$ then the value of $\frac{x^2-x y+y^2}{x^2+x y+y^2}=$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile