If xyz ≠ 0, xy 0, and x2 + xy + xz

Question

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

ccooley · Accepted Answer

Fun problem! The key with these is to do as much work as you can using logic and simplification, before you start testing cases. That's especially true on this problem, because there are three variables. If there are more than two variables in a problem, I usually try to avoid testing cases as much as possible - it just takes too long to consider all of the possibilities. The first thing I noticed is that x and y have the same sign . I also noticed that the inequality, x^2 + xy + xz < 0 , can be simplified like this: x(x+y+z) < 0 . This means that either x is negative, or x+y+z is negative, but not both. Now, go look at options I, II, and III. Option II stood out to me the most, because I already know something about x+y+z from my reasoning above. It doesn't HAVE to be negative, since the alternative is that x could just be negative. Then, if y and z were big enough, x+y+z would be positive. Eliminate B, D, and E , because they include option II. This would be a good point to guess! Or, we could keep going if we have enough time. Notice that the difference between (A) and (C) is that (A) doesn't include III, while (C) does. They both include I! So, at this point, we know that I has to be true without testing it. We only have to worry about III. That's a good reason to use the process of elimination! III says that if x < 0, z > 0. Well, if x < 0, from the above reasoning, we know that x + y + z > 0. We also know that x and y have the same sign. So, here's what we have: If x is negative, y is also negative, but x + y + z is positive . This should only happen if z is positive! You can't add three negative numbers together and get a positive result. So, z definitely has to be positive. III is true. Eliminate (A) and choose (C) .

KanishkM · Answer

IMO C xy > 0, and x^2 + xy + xz < 0 xy > 0, and x (x + y + z) < 0 Some cases before marking the options Case 1, x = -ive and y = -ive & For x (x+y+z) < 0, z > 0 and z > x + y Case 2, x = + ive and y = + ive & For x (x+y+z) < 0, z > -(x + y) III. If x < 0, then z > 0 This will be definitely true, case 1 holds good. I. x(y + z) < 0 True when it is Case 1, True when it is Case 2 II. x + y + z < 0 For case 1, this is not true For case 2, this is true

Nums99 · Answer

Can someone please simplify why it cannot be E

Aren't I II III independent statements?
If yes can we make use of the 3 of them to imply whether x is negative or no?

otautahi · Answer

Conditions: a) xyz ≠ 0 b) xy > 0 (it means xy is either both positive or both negative) c) x^2 + xy + xz < 0 (it means x^2 positive, xy must be positive from the second condition, and xz has to be negative) I. x(y + z) < 0, => xy + xz <0, from condition c, we can see xz is negative and greater than positive x^2 + xy combined, so this statement is true II. x + y + z < 0, we can not say this true. Maybe, maybe not. III. If x < 0, then z > 0. Here given x is negative. Condition c: x^2 + xy + xz < 0 x(x+y+z) <0, now to make x(x+y+z) negative/less than 0, we will need to make (x+y+z) positive. From condition b, we already know when x is negative, y has to be negative. So to make (x+y+z) positive, z has to be positive since x and y both are negative. So this statement is true. C) I and III only

bumpbot · Answer

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If xyz ≠ 0, xy > 0, and x2 + xy + xz < 0, then which of the following

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