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
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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.
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