for x^2>1 I would follow
x>|1| ==> x>1 and x>-1, since -1<x<=1 is not the solution area for x^2>1 we count only x>1
x>1 is the solution for x^2>1
x^2 > 1
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goalevan
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First method:
Whenever the form xy > 0 is seen, this can be interpreted as "x and y are the same sign" or "x and y are both positive or are both negative".
In the case of (x-1)(x+1) > 0, either both the quantities x-1 and x+1 are positive or they are both negative, so we have to consider both cases:
Positive case: x - 1 > 0 AND x + 1 > 0, so x > 1 AND x > -1. x > 1 holds since it is more restrictive.
Negative case: x - 1 < 0 AND x + 1 < 0, so x < 1 AND x < -1. x < -1 holds since it is more restrictive.
From the two cases combined, we have x > 1 OR x < -1.
Second method:
The form x^2 > k^2, where k is some positive number, can be interpreted as |x| > k, or x > k OR x < -k. This simply says that the distance of x from 0 is greater than k.
Both your methods are correct, just different interpretations.
A third method:
Think of the graph of x^2 and how it looks. It will be centered on the y-axis with the point (0,0), decreasing from negative infinity to x=0, and increasing from x=0 to infinity. Think about where this function is greater than 1. It is less than 1 only between -1 and 1, so outside of this range this inequality will hold true.
Whenever the form xy > 0 is seen, this can be interpreted as "x and y are the same sign" or "x and y are both positive or are both negative".
In the case of (x-1)(x+1) > 0, either both the quantities x-1 and x+1 are positive or they are both negative, so we have to consider both cases:
Positive case: x - 1 > 0 AND x + 1 > 0, so x > 1 AND x > -1. x > 1 holds since it is more restrictive.
Negative case: x - 1 < 0 AND x + 1 < 0, so x < 1 AND x < -1. x < -1 holds since it is more restrictive.
From the two cases combined, we have x > 1 OR x < -1.
Second method:
The form x^2 > k^2, where k is some positive number, can be interpreted as |x| > k, or x > k OR x < -k. This simply says that the distance of x from 0 is greater than k.
Both your methods are correct, just different interpretations.
A third method:
Think of the graph of x^2 and how it looks. It will be centered on the y-axis with the point (0,0), decreasing from negative infinity to x=0, and increasing from x=0 to infinity. Think about where this function is greater than 1. It is less than 1 only between -1 and 1, so outside of this range this inequality will hold true.
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pemdas
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a bit tricky, as the functional relationship is missing. Instead of y=x^2 we have x^2>1
any solution x>1 will hold true for -x<-1 and this outscores the interval (-1;1) too
Can we use y=x^2 here? A function must be continuous, and here it's not continuous on the interval (-1;1). I guess we will not use y=x^2 along with the last method described.
any solution x>1 will hold true for -x<-1 and this outscores the interval (-1;1) too
Can we use y=x^2 here? A function must be continuous, and here it's not continuous on the interval (-1;1). I guess we will not use y=x^2 along with the last method described.
goalevan wrote: A third method:
Think of the graph of x^2 and how it looks. It will be centered on the y-axis with the point (0,0), decreasing from negative infinity to x=0, and increasing from x=0 to infinity. Think about where this function is greater than 1. It is less than 1 only between -1 and 1, so outside of this range this inequality will hold true.
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pemdas
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@goalevan, I am not sure if GMAT is asking for two functions here.
Moreover, const. function f(x)=1 and f(x)=x^2 should produce the line and the parabola which opens upwards (x^2 positive coefficient, i.e. 1).
If we consider x^2>1 on the left and right-hand sides, then we need to reconsider the function f(x)=1 and convert this to f(x)>1 along the y-abscess, agree?
Perhaps the only way to navigate two functions here would be to draw one vertical asymptote to the graph of a function f(x)=x^2 on the right-hand side.
Moreover, const. function f(x)=1 and f(x)=x^2 should produce the line and the parabola which opens upwards (x^2 positive coefficient, i.e. 1).
If we consider x^2>1 on the left and right-hand sides, then we need to reconsider the function f(x)=1 and convert this to f(x)>1 along the y-abscess, agree?
Perhaps the only way to navigate two functions here would be to draw one vertical asymptote to the graph of a function f(x)=x^2 on the right-hand side.
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[email protected]
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Whatever goalevan has written is absolutely correct. But with one variation.
There were actually 2 possibilities. (i) as she said x > 1 and x < -1
and Second (ii) as -1 < x < 1
Now try putting as many values as you can and try negating any one of the option...
you will find that when the second case is there it does not suffice the original equation x^2 > 1.
Hence only the 1st case is correct. Thats it.
Hope this helped... Remember it is not enough to just solve an inequality, putting values is also very important. GMAT test on everything...
There were actually 2 possibilities. (i) as she said x > 1 and x < -1
and Second (ii) as -1 < x < 1
Now try putting as many values as you can and try negating any one of the option...
you will find that when the second case is there it does not suffice the original equation x^2 > 1.
Hence only the 1st case is correct. Thats it.
Hope this helped... Remember it is not enough to just solve an inequality, putting values is also very important. GMAT test on everything...
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GMATGuruNY
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No need to plug in more than 3 values.[email protected] wrote:Whatever goalevan has written is absolutely correct. But with one variation.
There were actually 2 possibilities. (i) as she said x > 1 and x < -1
and Second (ii) as -1 < x < 1
Now try putting as many values as you can and try negating any one of the option....
x² > 1.
x² - 1 > 0
(x+1)(x-1) > 0.
The CRITICAL POINTS are x=-1 and x=1.
These are the only values where the lefthand side is equal to 0.
To determine the range of x, plug in one value to the left and right of each critical point.
Thus, to determine where x²>1:
Plug in ONE value less than -1, ONE value between -1 and 1, and ONE value greater than 1.
If x=-2, then x² > 1. Thus, x<-1 is part of the range.
If x=0, then x² < 1. Thus, -1<x<1 is NOT part of the range.
If x=2, then x² > 1. Thus, x>1 is part of the range.
Thus, the ranges that satisfy x²>1 are x<-1 and x>1.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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