Question : If $(a-2)+\frac{1}{(a+2)} =-1$, the value of $\left (a+2 \right )^{2}+\frac{1}{(a+2)^{2}}$ is:
Option 1: 7
Option 2: 11
Option 3: 23
Option 4: 27
New: SSC CGL 2025 Tier-1 Result
Latest: SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 7
Solution :
Given:
$(a-2)+\frac{1}{(a+2)} =-1$
⇒ $(a+2)+\frac{1}{(a+2)}=3$ (Adding 4 on both sides)
Square both sides and use the formula, $(a+b)^2=a^2+b^2+2ab$
⇒ $(a+2)^2+(\frac{1}{(a+2)})^2+2(a+2)(\frac{1}{(a+2)})=9$
⇒ $(a+2)^2+(\frac{1}{(a+2)})^2+2=9$
⇒$(a+2)^2+\frac{1}{(a+2)^2}=9-2$
⇒$(a+2)^2+\frac{1}{(a+2)^2}=7$
Hence, the correct answer is 7.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
