Question : If $x$,$y$ and $z$ are real numbers such that $(x-3)^2+(y-4)^2+(z-5)^2=0$, then $(x+y+z)$ is equal to:
Option 1: –12
Option 2: 0
Option 3: 8
Option 4: 12
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 12
Solution : Given: $x,y$ and $z$ are real numbers such that $(x-3)^2+(y-4)^2+(z-5)^2=0$ We know, The addition of three perfect squares can be zero only when the individual numbers are zero. So, $(x-3)^2=0, (y-4)^2=0$ and $(z-5)^2=0$ ⇒ $x = 3, y = 4$ and $z = 5$. Then, $(x+y+z)$ = 3 + 4 + 5 = 12 Hence, the correct answer is 12.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : If $x+y+z=0$ and $x^2+y^2+z^2=40$, then what is the value of $x y+y z+z x?$
Question : If $(x-5)^2+(y-2)^2+(z-9)^2=0$, then value of $(x+y-z)$ is:
Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Question : $x,y,$ and $z$ are real numbers. If $x^3+y^3+z^3 = 13, x+y+z = 1$ and $xyz=1$, then what is the value of $xy+yz+zx$?
Question : If $x+y+z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1,$ and $xyz=-1$, then $x^3+y^3+z^3 $ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile