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What is the range of solutions for |x^2 - 4| > 3x ? 05 Nov 2014, 11:29
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What is the range of solutions for |x^2 - 4| > 3x ?

When I solved the equation, I got these roots:
(x-4)(x+1) > 0 and (x+4)(x-1) >> inequalities-trick-91482.html#p700500

Thanks!



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What is the range of solutions for |x^2 - 4| > 3x ? 06 Nov 2014, 06:40
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What is the range of solutions for |x^2 - 4| > 3x ?

When I solved the equation, I got these roots:
(x-4)(x+1) > 0 and (x+4)(x-1) < 0

Now, how can I plot this on the number line like this --->>> inequalities-trick-91482.html#p700500

Thanks!
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First check this: solving-quadratic-inequalities-graphic-approach-170528.html

What is the range of solutions for |x^2 - 4| > 3x ?

\(x^2 - 4 < 0\), when \(-2 < x < 2\). Hence, when \(-2 < x < 2\), \(|x^2 - 4| = -(x^2 - 4)\). So, in this case we would have \(-(x^2 - 4) > 3x\) --> \(x^2 + 3x -4 < 0\) --> \((x + 4)(x - 1) < 0\) --> \(-4<x<1\). Since we are in the range \(-2 < x < 2\), then we'd have \(-2 < x < 1\).

\(x^2 - 4 \geq{ 0}\), when \(x \leq{ -2}\) and \(x \geq{2}\). Hence, when \(x \leq{ -2}\) or \(x \geq{2}\), \(|x^2 - 4| = x^2 - 4\). So, in this case we would have \(x^2 - 4 > 3x\) --> \(x^2 - 3x -4 > 0\) --> \((x+1)(x-4)>0\) --> \(x<-1\) or \(x>4\). Since we are in the range \(x \leq{ -2}\) and \(x \geq{2}\), then we'd have \(x \leq{ -2}\) and \(x > 4\).

So, the the ranges of solutions for |x^2 - 4| > 3x, are \(x \leq{ -2}\), \(-2 < x < 1\), and \(x > 4\). We can combine first two ranges and we'll get \(x < 1\) and \(x > 4\).

Hope it helps.
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What is the range of solutions for |x^2 - 4| > 3x ? 06 Nov 2014, 07:03
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You could also graph the equations \(y=|x^2-4|\) and \(y=3x\) as I show in the attached image, find their intersections and identify the region where \(|x^2-4|>3x\). As you can see in the image, this happens whenever \(x<1\) or \(x>4\).

Cheers,
Dabral
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inequality-graph.png [ 18.99 KiB | Viewed 1921 times ]

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What is the range of solutions for |x^2 - 4| > 3x ? 06 Nov 2014, 09:24
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Bunuel and Dabral - Thank you both so much for pitching in with your answers. As much as I love knowing multiple solutions, much to my frustration, the inequality concept (for that matter a sizable chunk of math topics is tough for a non-engineer like me) escapes me and when it involves absolute operation, it only gets the best of me. Anywho, please look at the inline attachment and let me know where I am missing it. The reason why I am asking you that is because, I think I fairly understood the method outlined here (inequalities-trick-91482.html#p700500) and now steering into a different direction will only make it worse for me :)

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What is the range of solutions for |x^2 - 4| > 3x ? 06 Nov 2014, 10:35
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When you remove the absolute value sign, say in the first example where you equate \(|x^2-4|=x^2-4\), we also have an additional condition of \(x^2-4>0\), which is equivalent to \(x>2\) or \(x<-2\). You have to superimpose this condition.

As an aside, I do want to point out that I haven't seen such an inequality on the real GMAT exam, and if I had to bet my money then I would say that this problem is not appropriate for the GMAT.

Cheers,
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What is the range of solutions for |x^2 - 4| > 3x ? 06 Nov 2014, 22:59
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[quote="dabral"] ... additional condition of \(x^2-4>0\)[quote]

I am sorry for being thick, but I still cannot fathom why I have to consider a "> 0" situation when I remove the modulus operator? Clearly, I am missing something here.

TIA.
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What is the range of solutions for |x^2 - 4| > 3x ? 08 Nov 2014, 12:01
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When we remove the absolute value condition and say that \(|x^2-3|=x^2-3\), we can only do so when \(x^2-3>0\). Similarly, we say when \(x^2-3<0\), we have \(|x^2-3|=-(x^2-3)\). This is based on the definition of the absolute value.
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What is the range of solutions for |x^2 - 4| > 3x ? 12 Nov 2014, 23:02
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What is the range of solutions for |x^2 - 4| > 3x ?

When I solved the equation, I got these roots:
(x-4)(x+1) > 0 and (x+4)(x-1) >> inequalities-trick-91482.html#p700500

Thanks!
Show more

Responding to a pm:
Right! So the issue with your solution is this:

You have made two quadratics to remove the absolute value:

x^2 - 4 > 3x

and

-(x^2 - 4) > 3x

Mind you, there is a condition attached to removing the absolute value sign.

When do you get |x^2 - 4| = x^2 - 4?
You get it when x^2 - 4 is positive i.e. x^2 - 4 > 0
i.e. when (x-2)(x+2) > 0 i.e. x > 2 or x 4

Similarly, when is |x^2 - 4| = - (x^2 - 4)?
When x^2 - 4 is negative i.e. x^2 - 4 < 0
i.e. when (x-2)(x+2) < 0
So -2 < x < 2
When you solved this inequality, you got x = -4, 1
Note that -4 does not fall in this range so it is not a valid root. You only get x = 1 as a valid root. So for negative values, x < 1

That's how you get the correct answer.

Takeaway:
Never forget the absolute value conditions of
|x| = x if x is positive
|x| = -x if x is negative
When you make two equations to remove the absolute value sign, remember the attached conditions.



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