Question : If $x+ \frac{1}{9x}=4$, then the value of $9x^2+ \frac{1}{9x^2}$ is:
Option 1: 140
Option 2: 142
Option 3: 144
Option 4: 146
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: 142
Solution : Given: $x+ \frac{1}{9x}=4$ Multiply by 3 on both sides of the above equation, we get, ⇒ $3x+ \frac{1}{3x}=12$ Squaring both sides of the above equation, we get, ⇒ $(3x+ \frac{1}{3x})^2=12^2$ ⇒ $9x^2+ \frac{1}{9x^2}+2=144$ ⇒ $9x^2+ \frac{1}{9x^2}=144–2$ ⇒ $9x^2+ \frac{1}{9x^2}=142$ Hence, the correct answer is 142.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x=2+\sqrt3$, then the value of $\frac{x^{2}-x+1}{x^{2}+x+1}$ is:
Question : If $x+\frac{1}{x}=1$, then the value of $\frac{x^2+7x+1}{x^2+11x+1}$:
Question : If $(x+\frac{1}{x})$ = 5, then the value of $\frac{5x}{x^{2}+5x+1}$ is:
Question : If $x+\frac{1}{x}=-14$, and $x<-1$, what will be the value of $x^2-\frac{1}{x^2}$?
Question : If $x+\frac{1}{x}=6$, then find the value of $\frac{3 x}{2 x^2-5 x+2}$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile