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dddanny2006
- Master | Next Rank: 500 Posts
- Posts: 209
- Joined: Thu Jan 12, 2012 12:59 pm
Is x ≥ 0?
1) x^2 = 9x
2) |x| = -x
Going back to the article,Statement 1 is absolutely sufficient.When it comes to statment 2,this is my way of solving it and it contradicts with the answer you've provided.
|x|=-x =>x=+(-x)=-x,this is true when x>0, and x=-(-x),this is true when x<0
Therefore statement 2 gives us 2 conditions x=x and x=-x.Now since x=-x when x>0,we cant test 0 there.Now,we also have x=x when x<0,so -5=-5,-6=-6,these dont satisfy the condition in the question stem and thus its a NO.Thus the answer is D in contradiction to A what the book says.
In which way is my understanding of the question wrong?
I have followed this technique from here--
https://www.manhattangmat.com/strategy-s ... -value.cfm
Absolute value expressions start to become difficult when variable expressions are placed inside the bars. For example, /x/. Upon a cursory examination, the expression /x/ seems like it should be equal to x. Since there is no sign in front of the x, the absolute value bars should be able to be removed without jeopardizing the "guarantee of positive." What this line of reasoning fails to account for, however, is that x itself could be negative! When dealing with absolute value expressions that contain variables, two scenarios must be considered: (1) the scenario whereby the expression inside the bars is positive and (2) the scenario whereby the expression inside the bars is negative.
In this example, for scenario (1) if x > 0, the expression /x/ can simply be represented as x; for scenario (2) if x < 0, the expression /x/ must be represented as (-x). Notice that in the negative scenario, we don't simply remove the absolute value bars. We remove the absolute value bars and negate the entire expression within.
Let's look at a more complicated example: the expression /x - 3/. As always, we must consider both the positive and negative scenarios. When is the expression inside the absolute value bars positive? Not simply when x > 0, but when x - 3 > 0 or when x > 3. Likewise the expression will be negative when x < 3.
To recap, the two scenarios are:
(1) /x - 3/ can be rewritten as x - 3 when x > 3
(2) /x - 3/ can be rewritten as -(x - 3) or 3 - x when x < 3
One more for the road: /3x + y/.
(1) /3x + y/ can be rewritten as 3x + y when 3x + y > 0
(2) /3x + y/ can be rewritten as -(3x + y) when 3x + y < 0
1) x^2 = 9x
2) |x| = -x
Going back to the article,Statement 1 is absolutely sufficient.When it comes to statment 2,this is my way of solving it and it contradicts with the answer you've provided.
|x|=-x =>x=+(-x)=-x,this is true when x>0, and x=-(-x),this is true when x<0
Therefore statement 2 gives us 2 conditions x=x and x=-x.Now since x=-x when x>0,we cant test 0 there.Now,we also have x=x when x<0,so -5=-5,-6=-6,these dont satisfy the condition in the question stem and thus its a NO.Thus the answer is D in contradiction to A what the book says.
In which way is my understanding of the question wrong?
I have followed this technique from here--
https://www.manhattangmat.com/strategy-s ... -value.cfm
Absolute value expressions start to become difficult when variable expressions are placed inside the bars. For example, /x/. Upon a cursory examination, the expression /x/ seems like it should be equal to x. Since there is no sign in front of the x, the absolute value bars should be able to be removed without jeopardizing the "guarantee of positive." What this line of reasoning fails to account for, however, is that x itself could be negative! When dealing with absolute value expressions that contain variables, two scenarios must be considered: (1) the scenario whereby the expression inside the bars is positive and (2) the scenario whereby the expression inside the bars is negative.
In this example, for scenario (1) if x > 0, the expression /x/ can simply be represented as x; for scenario (2) if x < 0, the expression /x/ must be represented as (-x). Notice that in the negative scenario, we don't simply remove the absolute value bars. We remove the absolute value bars and negate the entire expression within.
Let's look at a more complicated example: the expression /x - 3/. As always, we must consider both the positive and negative scenarios. When is the expression inside the absolute value bars positive? Not simply when x > 0, but when x - 3 > 0 or when x > 3. Likewise the expression will be negative when x < 3.
To recap, the two scenarios are:
(1) /x - 3/ can be rewritten as x - 3 when x > 3
(2) /x - 3/ can be rewritten as -(x - 3) or 3 - x when x < 3
One more for the road: /3x + y/.
(1) /3x + y/ can be rewritten as 3x + y when 3x + y > 0
(2) /3x + y/ can be rewritten as -(3x + y) when 3x + y < 0
