inequality

[Randyyy]

Hey, the given inequality is as follows:
(4-x²)²>(x²-4x+4)²
I begin with factoring both sides of the inequality as follows:
(-2+x)²(2+x)²>(x-2)⁴
Here is where I run into an issue. If I divide by (x-2)² and get the following:
(2+x)²>(x-2)² I lose a solution because solving this now gives me the solution of x>0.
If I instead divide the entire RHS (2+x)²/(x-2)²>1 It yields the correct solutions 0<x<2 II x > 2
What is it I am overlooking?

 

[Steven G]

While I try to follow what you wrote can you explain what II means in 0<x<2 II x > 2

You can use any symbols you like but you need to define them.

 

[Steven G]

Your mistake is that you may be dividing by 0? In your 1st method, you would be dividing by 0 if x=2. You need to check if x=2 is a solution and it is!

Method 2: You incorrectly replaced (x-2)^2/(x-2)^2 with 1. This is not always true! If x=2 then you replacing 1 for (x-2)^2/(x-2)^2 was wrong. Replacing an expression with a wrong value will yield an error.

Rather than multiply or dividing by an expression that might be 0 you should add or subtract that term from both sides.

This will give you (-2+x)²(2+x)²-(x-2)⁴ >0

You should continue from the line above. I would factor first.

One last comment. In your case what you divided by could never be negative but sometimes they could be. For example if you multiplied (or divided) both sides of an inequality by x-2 then you would not know if you should change the direction of the inequality sign since x-2 can be negative.

 

[Randyyy]

What I really meant to state was that the solutions is 0<x<2 or x>2 and used "II" to represent or. My mistake, I should have been more clear with what I meant.
So before I divide by (x-2)^2 (if I chose to) I would have to check if x=2 is a solution to my original inequality which in this case is true and hence I make the mistake of dividing by zero. But why does it work if I divide the entire RHS and what should be left if I divide the entire RHS if not 1? Or is it just that I get lucky that it yields the correct solution when I divide the way I did? Because I get that if it was negative the signs would have been flipped but I am dividing by a square so it shouldn´t be able to be negative (although it could be 0 just like you said).

Following your suggestion I do the following:
(-2+x)²((2+x)²-(-2+x)²)>0
(-2+x)²*8x>0

8x=0 or (-2+x)=0
giving the solutions that 0<x<2 and x>2

 

[pka]

Hey, the given inequality is as follows: (4-x²)²>(x²-4x+4)²
I begin with factoring both sides of the inequality as follows:
(-2+x)²(2+x)²>(x-2)⁴
Here is where I run into an issue. If I divide by (x-2)² and get the following:
(2+x)²>(x-2)² I lose a solution because solving this now gives me the solution of x>0.
If I instead divide the entire RHS (2+x)²/(x-2)²>1 It yields the correct solutions 0<x<2 II x > 2

Do you realize that \((4-x^2)^2=(x^2-4)^2=(x-2)^2(x+2)^2~?\)

 

[Randyyy]

Do you realize that \((4-x^2)^2=(x^2-4)^2=(x-2)^2(x+2)^2~?\)

Yes, I agree. That is what I factored (4-x²)² to be though. I just have it written as (2+x) instead of (x+2) etc because a^2-b^2 so I wrote it in that order to make sure I didn´t mess up the order of the terms inside the parenthesis. But it would probably look better if I wrote it as you did rather than the way I have it written currently.

 

[Steven G]

What I really meant to state was that the solutions is 0<x<2 or x>2 and used "II" to represent or. My mistake, I should have been more clear with what I meant.
So before I divide by (x-2)^2 (if I chose to) I would have to check if x=2 is a solution to my original inequality which in this case is true and hence I make the mistake of dividing by zero.

If x=2 is the only solution and you divided by x-2 then yes you made a mistake by dividing by 0.
Once you realize that x=2 works or doesn't work (If it works then remember to include it in your solution and if it doesn't work then exclude it from your answer) you can assume x is NOT 2. Why? We already considered that case! Then you proceed as you did. Now in the end you got that x>0 and x is not 2. But you know that x=2 does work. Combining yields that x>0.

 

[Randyyy]

Thanks for the help! I think I get the premise of the error I made. So if possible I should always try to subtract or add and factorise rather than dividing and multiplying with terms that could either be 0 or could be negative.

 

[Eduleo]

···
So before I divide by (x-2)^2 (if I chose to) I would have to check if x=2 is a solution to my original inequality which in this case is true and hence
···

I can not see that [math]x=2[/math] is a solution of the original inequality [math](4-x^2)^2>(x^2-4x+4)^2[/math]. For [math]x=2[/math] we get [math]0>0[/math] so, it is not a solution... isn't?

 

[Duplicity]

I can not see that [math]x=2[/math] is a solution of the original inequality [math](4-x^2)^2>(x^2-4x+4)^2[/math]. For [math]x=2[/math] we get [math]0>0[/math] so, it is not a solution... isn't?

Actually I was thinking the same way. Why in some answers, people say “x=2 works” ? It surely does not work.

 

[Eduleo]

The curious thing is that, after dividing both sides by [math](x-2)^2[/math], then [math]x=2[/math] is now a solution for the resulting inequality: [math](2+x)^2>(x-2)^2[/math]. So it seems that there is a problem with that operation.... we have to take into account that (from original inequality, x is not equal to 2. So combining both restult (the one obtainded from this new inequality (x>0) and the original one (x not equal 2) we get the final result: [math]x>0[/math] and x not equal 2.

 

[Randyyy]

I can not see that [math]x=2[/math] is a solution of the original inequality [math](4-x^2)^2>(x^2-4x+4)^2[/math]. For [math]x=2[/math] we get [math]0>0[/math] so, it is not a solution... isn't?

Hey, what I meant is that it is a solution to the equation but it does not satisfy the inequality because like you said, 0>0 is not true.

Actually I was thinking the same way. Why in some answers, people say “x=2 works” ? It surely does not work.

I think what people mean with "it works" is that it is a critical value but in our case it does not satisfy the inequality.

or they are referring to the fact it is A solution that does satisfy the inequality after I make the error dividing.

 

[Deleted member 4993]

What I really meant to state was that the solutions is 0<x<2 or x>2 and used "II" to represent or. My mistake, I should have been more clear with what I meant.
So before I divide by (x-2)^2 (if I chose to) I would have to check if x=2 is a solution to my original inequality which in this case is true and hence I make the mistake of dividing by zero. But why does it work if I divide the entire RHS and what should be left if I divide the entire RHS if not 1? Or is it just that I get lucky that it yields the correct solution when I divide the way I did? Because I get that if it was negative the signs would have been flipped but I am dividing by a square so it shouldn´t be able to be negative (although it could be 0 just like you said).

Following your suggestion I do the following:
(-2+x)²((2+x)²-(-2+x)²)>0
(-2+x)²*8x>0

8x=0 or (-2+x)=0
giving the solutions that 0<x<2 and x>2

You correctly got:

(-2+x)²*8x > 0 \(\displaystyle \ \ \to \ \ \) (-2+x)² * x > 0

on the LeftHandSide

(-2+x)² is positive when \(\displaystyle x \ne 2\)

so:

(-2+x)² * 8x is positive when x > 0 AND \(\displaystyle x \ne 2\)

 

[Steven G]

I can not see that [math]x=2[/math] is a solution of the original inequality [math](4-x^2)^2>(x^2-4x+4)^2[/math]. For [math]x=2[/math] we get [math]0>0[/math] so, it is not a solution... isn't?

Oops, x=2 is not a solution

 

[Steven G]

Actually I was thinking the same way. Why in some answers, people say “x=2 works” ? It surely does not work.

They shouldn't and if they do they are wrong. I was wrong to say x=2 is a solution!

 

[Steven G]

The curious thing is that, after dividing both sides by [math](x-2)^2[/math], then [math]x=2[/math] is now a solution for the resulting inequality: [math](2+x)^2>(x-2)^2[/math]. So it seems that there is a problem with that operation.... we have to take into account that (from original inequality, x is not equal to 2. So combining both restult (the one obtainded from this new inequality (x>0) and the original one (x not equal 2) we get the final result: [math]x>0[/math] and x not equal 2.

It is not curious or strange that x=2 is a solution to the last inequality. The reason is that the OP reduced (x-2)/(x-2) to 1 which is not true if x=2. So in the end if x is 2, then you made a mistake by replacing (x-2)/(x-2) with 1. If for example if you replaced (x-2)/(x-2) with 7 then of course you will arrive at a wrong answer. The wrong answer can be x=2 just like any other number.

 

[joshuaa]

this is a very simple inequality

you don't need all this complication

since [MATH](4 - x^2)^2 = (x^2 - 4)^2[/MATH], you just need to solve the inequality [MATH]2[/MATH] times

one time
[MATH]4 - x^2 > x^2 - 4x + 4[/MATH]
And another time
[MATH]x^2 - 4 > x^2 - 4x + 4[/MATH]

 

[HallsofIvy]

I would not say that "x= 2" was a solution to the original inequality. However, since this is a polynomial, so a continuous function, places where "p(x)> q(x)" and "p(x)< q(x)" are always separated by places where "p(x)= q(x)". Often the best way to solve an inequality is to first solve the associated equation.

Here, the inequality is \(\displaystyle (4-x^2)^2>(x^2-4x+4)^2\). The "associated equarion" is \(\displaystyle (4- x^2)^2= (x^2- 4x+ 4)^2\). That is the same as \(\displaystyle (2- x)^2(2+ x)^2= (x- 2)^4\). It is obvious that x= 2 is a solution. If x is not 2 we can divide both sides by \(\displaystyle (x- 2)^2\) to get \(\displaystyle (2+ x)^2= (2- x)^2\). That is the same as \(\displaystyle 4+ 4x+ x^2= x^2- 4x+ 4\) which reduces to \(\displaystyle 4x= -4x\) and then to \(\displaystyle x= 0\). That is, \(\displaystyle (4- x^2)^2= (x^2- 4x+ 4)^2\) at x= 0 and x= 2.

That means that the inequality may be satisfied in three intervals, \(\displaystyle -\infty< x< 0\), \(\displaystyle 0< x< 2\), and \(\displaystyle 2< x< \infty\). Now we just need to check the inequality at one x value in each of those intervals.

For \(\displaystyle x< 0\) take x= -1. \(\displaystyle (4- (-1)^2)^2= 3^2= 9\) and \(\displaystyle ((-1)^2- 4(-1)+ 4)^2= (1+ 4+ 4)^2= 9^2= 81\). 9< 81 so the inequality is false for all x< 0.

For \(\displaystyle 0< x< 2\) take x= 1. \(\displaystyle (4- 1^2)^2= 3^2= 9\) and \(\displaystyle ((1)^2- 4(1)+ 4)^2= 1^2= 1\). 9> 1 so the inequality is true for all 0< x< 2.

Finally, for \(\displaystyle x> 2\) take x= 3. \(\displaystyle (4- 3^2)^2= (-5)^2= 25\) and \(\displaystyle (3^2- 4(3)+ 4)^2= (9- 12+ 4)^2= 1\) 25> 1 so the inequality is true for all x>2.

\(\displaystyle (4-x^2)^2>(x^2-4x+4)^2\) is true for \(\displaystyle (0. 2)\cup (2, \infty)\), all positive x except 2.