18 Views
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)
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.
SSC CGL Complete Guide
Candidates can download this ebook to know all about SSC CGL.
Download EBookKnow More About
Related Questions
Upcoming Exams
Admit Card Date:
5 Jun, 2025
- 15 Jun, 2025
Application Date:
1 Jun, 2025
- 5 Jun, 2025
1st Stage CBT
Exam Date:
5 Jun, 2025
- 24 Jun, 2025
Application Date:
3 Jun, 2025
- 23 Jun, 2025
Application Date:
22 Apr, 2025
- 5 Jun, 2025