Question : If $(a+1)^2+(a+2)^2=16$, then what is the value of $40+12 a+4 a^2 ?$
Option 1: 56
Option 2: 62
Option 3: 74
Option 4: 52
New: SSC CGL 2025 Tier-1 Result
Latest: SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 62
Solution :
Given: $(a+1)^2+(a+2)^2=16$
⇒ $a^2+2a+1+a^2+4a+4 = 16$
⇒ $2a^2+6a=11$ -------------(i)
Now, $40+12 a+4 a^2$
= $40+2(2a^2+6a)$
= $40+22$
= 62
Hence, the correct answer is 62.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates BrochureYour Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.




