need help on solving Irrational modular inequalities

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Broker209

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greetings.
im in need of help solving the exercise attached to this post the exercise is about Irrational modular inequalities ,its a new topic im being given at highschool and i have no clue on how to solve it, pls any help would be appreciated.
thx
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Broker209 said:
no. i was only taugth taught to solve it by finding the domain and values of x
Click to expand...
OK, good. So what have you tried to find the domain? Did you try factoring the denominator. Is the numerator a perfect square or factorable.

To receive help we need to know what you have tried and where you are stuck. So please follow our guidelines (you did read them, correct?) and show us you work, even if you know it is wrong.
 
Here is a hint. I would rewrite the EQUIVALENT of the problem without the absolute value bars which will give you a few cases to consider.
For the record, I do not think that this will be easy to solve for you. When you break this problem up into cases you need to be very carely.
Just remember that |x| = x only when x> 0 and -x otherwise.
 
To start you off:

You are going to need to look at some critical points. For example, when is the number under the radical negative? zero? positive? (In other words you need to solve [math]2 - x - x^2 = 0[/math]. Obviously if [math]2 - x - x^2 < 0[/math] we don't have a solution.) Another set of critical points will be when is the number inside the absolute value brackets negative? zero? positive?

These are the points you need to test.

-Dan
 
Broker209 said:
im in need of help solving the exercise attached to this post the exercise is about Irrational modular inequalities ,its a new topic im being given at highschool and i have no clue on how to solve it, pls any help would be appreciated.
Click to expand...
The problem: \(\displaystyle \dfrac{\sqrt{2-x-x^2}}{|x^2-1|+|x-1|}>1\)
Note that if \(\displaystyle x=1\) you have \(\displaystyle 0/0\). If \(\displaystyle x\ne 1\) the denominator is positive.
Also see that you must have \(\displaystyle 2-x-x^2=(2+x)(1-x)>0\) for the radical to be defined.
I like the idea of looking at the graph.
 
pka said:
The problem: \(\displaystyle \dfrac{\sqrt{2-x-x^2}}{|x^2-1|+|x-1|}>1\)
Note that if \(\displaystyle x=1\) you have \(\displaystyle 0/0\). If \(\displaystyle x\ne 1\) the denominator is positive.
Also see that you must have \(\displaystyle 2-x-x^2=(2+x)(1-x)>0\) for the radical to be defined.
I like the idea of looking at the graph.
Click to expand...
thank you
 
t
topsquark said:
To start you off:

You are going to need to look at some critical points. For example, when is the number under the radical negative? zero? positive? (In other words you need to solve [math]2 - x - x^2 = 0[/math]. Obviously if [math]2 - x - x^2 < 0[/math] we don't have a solution.) Another set of critical points will be when is the number inside the absolute value brackets negative? zero? positive?

These are the points you need to test.

-Dan
Click to expand...
thank you
 
I am probably wrong but I disagree with the advice given. How will the critical points help if the right hand side is 1?
 
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