Quadratic inequality involving Modulus Function

  • Thread starter LiHJ
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  • #1
LiHJ
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Homework Statement


Dear Mentors and PF helpers,

I saw this question on a book but couldn't understand one part of it.

Here the question:
Solve the following inequality
I copied the solution as below


Homework Equations



image.jpg

The Attempt at a Solution


I don't understand why the numerator in step (1) can become 1 in step (2).
Can someone explain the concept behind.
Thank you for your time
 

Answers and Replies

  • #2
ehild
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Homework Statement


3. The Attempt at a Solution [/B]
I don't understand why the numerator in step (1) can become 1 in step (2).
Can someone explain the concept behind.
Thank you for your time
The numerator at 1) is (x-1)2, EdiT: (x-1)2 + 1, it can not be negative. You can divide by it, and the inequality holds.
 
Last edited:
  • #3
PWiz
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The numerator at 1) is (x-1)2, it can not be negative. You can divide by it, and the inequality holds.
I don't think ##x^2 -2x +2## factorizes to ##(x-1)^2## , and since the discriminant of the numerator in step 1 is negative, there will be no real solution to it. Hence, the numerator can be taken out of the equation by cross multiplication, and this leads to step 2. (I think so)
 
  • #4
ehild
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I don't think ##x^2 -2x +2## factorizes to ##(x-1)^2## , and since the discriminant of the numerator in step 1 is negative, there will be no real solution to it. Hence, the numerator can be taken out of the equation by cross multiplication, and this leads to step 2. (I think so)
You are right, I made a mistake when typing in. The numerator is (x-1)2+1, can not be negative.
If the numerator could not be zero, (had no roots) it could be entire negative. Dividing by it would change the inequality then . 2) follows from 1) only when the numerator is always positive.
 
  • #5
PWiz
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Just curious - is step 4) unnecessary? Can step 3 not be simplified to ##x^2-2 |x|## < 0 (Since 1/a can never be equal to 0)?
 
  • #6
ehild
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Just curious - is step 4) unnecessary? Can step 3 not be simplified to ##x^2-2 |x|## < 0 (Since 1/a can never be equal to 0)?[/QUOTE ]

You are right . From ##\frac {1}{x^2-|x|}<0## (equality is not allowed) x^2-|x|<0 follows. Step 4) is not needed. So |x|(|x|-2)<0 --> |x|<2.
 
  • #7
LiHJ
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Dear ehild and PWiz,

I have corrected it.

image.jpg


Please help me to do corrections if there's error.
There's one more thing that I need clarification is that on ehild 2nd post, states that if the numerator could not be zero, it coluld be entire negative.

What this means.
I know that if we multiply or divide negatives, the sign of inequality would change.
 
  • #8
ehild
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The numerator is always positive. So we need to drop the equality.

You can not divide a non-zero number by zero, so the denominator can not be zero, either.

You can divide the inequality by the positive numerator, so the inequality simplifies to the form 1/A<0. Step 4) is not necessary. You know that 1/A is negative only when A is negative. But you may do 4), it is correct. And the whole solution is correct, but |x|<0 is impossible,so exlude x=0 from he solution.

When I said that if the numerator can not be zero, it can be either entirely positive or entirely negative, it was answer to PWiz.
 
Last edited:
  • #9
PWiz
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As ehild has correctly pointed out, if the left side of the equation cannot equal 0, then the inequality can be simplified to being less than 0 as you've done is step 3(the expression on the LHS is -ve). For step 5, you can avoid the inequality |x|<0 by dividing both sides by |x| right at the beginning of step 5, and the inequality will be preserved (since |x| is always positive). So you get |x|-2 <0 (only), and then you can get the proper solution, just as how you've done in the last step.
 
Last edited:
  • #10
ehild
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X=0 has to be excluded from the solution.
 
  • #11
LiHJ
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Thank you ehild and PWiz,

Now I changed the question the other way round, it has to be less than or equal to zero.

image.jpg
 
  • #12
ehild
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A product of two factors can be positive only when either both factors are positive or both of them are negative. One of the factors is |x|. It is never negative, so the other factor |x|-2 is also positive.
 

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