Question : If $x^{2}+y^{2}+z^{2}=14$ and $xy+yz+zx=11$, then the value of $(x+y+z)^{2}$ is:
Option 1: 16
Option 2: 25
Option 3: 36
Option 4: 49
Answer (1)
23rd Jan, 2024
Correct Answer: 36
Solution :
Given: $x^{2}+y^{2}+z^{2}=14$ and $xy+yz+zx=11$
Thus, $(x+y+z)^{2}=x^2+y^2+z^2+2(xy+yz+zx)$
Putting the values, we get
$(x+y+z)^{2}=14+2(11)$
⇒ $(x+y+z)^{2}=14+22$
⇒ $(x+y+z)^{2}=36$
Hence, the correct answer is 36.
Comments (0)
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Get Updates Brochure
Thank You!
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Application Date:
23 Oct, 2024
- 25 Dec, 2024
Application Date:
9 Dec, 2024
- 30 Dec, 2024
Application Date:
2 Dec, 2024
- 31 Dec, 2024
Exam Date:
3 Jan, 2025
- 16 Jan, 2025
General Insurance Corporation Of India Assistant Manager Exam
- (GIC ASSISTANT MANAGER SCALE I EXAM)
Exam Date:
5 Jan, 2025
- 5 Jan, 2025
