If x2 + y2 + z2 - xy - yz - zx < 0 then which one statement is correct.

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03 Nov 2009 Which one statement is correct

If x2 + y2 + z2 - xy - yz - zx < 0 then which one statement is correct.

OPtion

1) x = y = z
2) x > y > z
3) x < y < z
4) x ≠ y = z
5) x = y ≠ z
6) x ≠ y ≠ z
7) x, y, z have negative values
8) any one of x, y, z is negative
9) any two of x, y, z are negative
10) Not possible

Solution

x2 + y2 + z2 – xy - yz – zx < 0
=> 2x2 + 2y2 + 2z2 – 2xy - 2yz – 2zx < 0
=> x2 + y2 - 2xy + y2 + z2 - 2yz + z2 + x2 - 2zx < 0
=> (x – y)2 + (y – z)2 + (z – x)2 < 0
(x – y)2 + (y – z)2 + (z – x)2 ,
This expression is the sum of three perfect square numbers and it is not possible to make this expression less than 0.

Correct Option: 10 Best Solution (5)

Anantha Krishna   14 years AGO

if x=y=z, then the equation becomes 0, in all other cases it will be more than 0. So it is not possible.

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SURAJ   14 years AGO

B'coz if x2 + y2 + z2 – xy - yz – zx < 0
then x2 + y2 + z2 < xy + yz + zx which is not possible

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vijaya   14 years AGO

as go on solving the equation for x,y,z values we get above statement is not possible.

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sujit   14 years AGO

x2 + y2 + z2 - xy - yz - xz < 0
=>x2 + y2 + z2 < xy + yz + xz
which is not possible

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Tijo Jose   14 years AGO

Put some arbitrary values for x, y, z and evaluate the expression. None of them seems to yield a negative answer.

For example,

1. Put x=y=z=1
2. Put x=3, y=2, z=1
3. Put x=1, y=2, z=3
4. Put x=1, y=z=2
5. Put x=y=1, z=2
6. Put x=1, y=2, z=3
7. Put x=-1, y=-2, z=-3
8. Put x=-1, y=2, z=3
9. Put x=-1, y=-2, z=3

Every-time, x2 + y2 + z2 - xy -yz - xz yields a non-negative number.

Hence, the result follows.

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