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If x, y and z are real numbers, then
x 2 + 4 y 2 + 9 z 2 – 6yz –3zx – 2xy is always
x 2 + 4 y 2 + 9 z 2 – 6yz –3zx – 2xy is always
- a)positive
- b)non -positive
- c)zero
- d)non-negative
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If x, y and z are real numbers, thenx 2 + 4 y 2 + 9 z 2 – 6yz &n...
x2 + 4y2 + 9z2 – 6yz – 3zx – 2xy
= x2 + (2y)2 + (3z)2 – (2y) (3z) – (3z)(x) – x (2 y) > 0
x, y, z
[∵ a2 + b2 + c2 - ab - bc - ca > 0]
= x2 + (2y)2 + (3z)2 – (2y) (3z) – (3z)(x) – x (2 y) > 0
x, y, z[∵ a2 + b2 + c2 - ab - bc - ca > 0]
Most Upvoted Answer
If x, y and z are real numbers, thenx 2 + 4 y 2 + 9 z 2 – 6yz &n...
Cannot be negative.
This is because the square of any real number is always non-negative (it is either zero or positive), so x^2, 4y^2, and 9z^2 are all non-negative. Therefore, their sum x^2 + 4y^2 + 9z^2 is also non-negative (it is zero only if all three terms are zero), and cannot be negative.
This is because the square of any real number is always non-negative (it is either zero or positive), so x^2, 4y^2, and 9z^2 are all non-negative. Therefore, their sum x^2 + 4y^2 + 9z^2 is also non-negative (it is zero only if all three terms are zero), and cannot be negative.
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If x, y and z are real numbers, thenx 2 + 4 y 2 + 9 z 2 – 6yz –3zx – 2xy is alwaysa)positiveb)non -positivec)zerod)non-negativeCorrect answer is option 'D'. Can you explain this answer?
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