Nums99 wrote:
Hi
VeritasKarishmaI had a doubt and its not clear to me yet.
If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?
I. x(y + z) < 0
II. x + y + z < 0
III. If x < 0, then z > 0.
A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II, and III
Can you please help me simplify why it cannot be E
xyz ≠ 0 implies none of x, y and z is 0.
xy > 0 implies both x and y have the same sign. Either both are positive or both negative.
x^2 + xy + xz < 0 implies xz is negative so x and z have opposite signs. x^2 is positive because squares cannot be negative. xy is positive we have already benn given so xz must be negative.
I. x(y + z) < 0
We know that the expression x^2 + xy + xz is negative
x^2 must be positive. So if we remove x^2 from above, the expression (xy + xz) will become even more negative.
Hence, this must be true.
II. x + y + z < 0
x^2 + xy + xz < 0
x(x + y + z) < 0
If x is positive, (x+y+z) < 0
If x is negative, (x + y + z) > 0 (Take x = -3, y = -1 and z = 100 for example)
So this is not necessarily true.
III. If x < 0, then z > 0.
Correct. x and z have opposite signs. So this must be true.
Answer (C)
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Karishma
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