Question
If x, y and z are real numbers, the x^ 2 +4 y^ 2 +9 z^ 2 -6 y z-3 z x-2 x y is always
BITSAT
Mathematics
Basic of Mathematics
2020
BITSAT 2020
BITSAT Mathematics Question (2020) — Solution
Options
- A. positive
- B. non-positive
- C. zero
- D. non-negative
Answer
D. non-negative
Step-by-step solution
x^ 2 +4 y^ 2 +9 z^ 2 -6 y z-3 z x-2 x y =x^ 2 +(2 y)^ 2 +(3 z)^ 2 -(2 y)(3 z)-(3 z)(x)- x(2 y) 0 x, y, z [ a ^ 2 + b ^ 2 + c ^ 2 - ab - bc - ca 0 ]
Related: Mathematics — Basic of Mathematics · All PYQ Banks