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
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
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 ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
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
Admit Card Date:
14 Nov, 2024
- 21 Nov, 2024
Tier II
Exam Date:
18 Nov, 2024
- 18 Nov, 2024
