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x^2 > 1

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by siddhans » Sun Jul 10, 2011 2:52 am
How to solve this ?? Please explain detailed steps?

Method1 :

x^2 > 1


x^2 - 1 > 0

=> (x-1)(x+1) > 0


What next ?

after this?

x > 1 or x < -1 ??? Do the '<' and '>' signs change???

I am confused






Method 2:

X ^2 > 1

x > 1 or x < -1


Are both the methods correct?? Where am i going wrong?
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by pemdas » Sun Jul 10, 2011 3:08 am
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
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by goalevan » Sun Jul 10, 2011 3:11 am
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.
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by pemdas » Sun Jul 10, 2011 3:41 am
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.
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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by goalevan » Sun Jul 10, 2011 9:35 am
We can represent each side of the inequality as a separate function:

f(x) = x^2
g(x) = 1

See this image:
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Last edited by goalevan on Sun Jul 10, 2011 1:05 pm, edited 1 time in total.
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by pemdas » Sun Jul 10, 2011 12:46 pm
@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.
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by goalevan » Sun Jul 10, 2011 1:04 pm
Take a look at the image, you can see that f(x) = x^2 exceeds g(x) = 1 for x < -1 and x > 1
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by [email protected] » Fri Mar 16, 2012 7:46 am
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...
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by GMATGuruNY » Fri Mar 16, 2012 10:02 am
[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....
No need to plug in more than 3 values.

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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