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kartik222
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X^2<x absolute value question 05 Oct 2012, 08:09
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Hi Bunuel,

What exactly is the solution for:

x^2 x(x-1)0 => x>1 ..OR.. x>0 , x-1 x 0x (signed changed)

I really appreciate your help.

-K



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X^2<x absolute value question 05 Oct 2012, 08:22
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kartik222
Hi Bunuel,

What exactly is the solution for:

x^2<x

From what I understand, it is:

x^2-x<0 => x(x-1)<0
either x<0 , x-1>0 => x>1 ..OR.. x>0 , x-1<0 => x<1, so there can be two possibilities here.
BUT saw somewhere that x^2<x ===> 0<x<1, is this true? according to me this is just one possibility.

In the similar way can u please also explain x^2>x (signed changed)

I really appreciate your help.

-K
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x^2<x --> x(x-1) < 0 --> roots are 0 and 1 --> "<" sign indicates that the solution lies between the roots: 0<x<1.

OR:
x(x-1) < 0
x<0 and x-1>0 (x>1), which is not possible, x cannot be simultaneously less than zero and more than 1.
x>0 and x-1<0 (x<1) --> 0<x<1.

As for x^2>x. x(x-1) > 0 --> roots are 0 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<0 or x>1.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
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X^2<x absolute value question 05 Oct 2012, 10:22
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thanks a lot bunuel!!

-K
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X^2<x absolute value question 07 Oct 2012, 21:43
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kartik222
Hi Bunuel,

What exactly is the solution for:

x^2<x

From what I understand, it is:

x^2-x<0 => x(x-1)<0
either x<0 , x-1>0 => x>1 ..OR.. x>0 , x-1<0 => x<1, so there can be two possibilities here.
BUT saw somewhere that x^2<x ===> 0<x<1, is this true? according to me this is just one possibility.

In the similar way can u please also explain x^2>x (signed changed)

I really appreciate your help.

-K
Show more

When solving such inequalities, keep in mind that it is not a good idea to cancel off variables.
So x^2-x<0 => x(x-1)<0
and x^2>x => x(x-1) > 0

Now, check out this post for details on how to quickly solve x(x-1)<0 or x(x-1)>0

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/06 ... e-factors/
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X^2<x absolute value question 08 Oct 2012, 20:01
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Thanks Karishma. Your article explained the concept clearly
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EvaJager
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X^2<x absolute value question 09 Oct 2012, 07:14
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kartik222
Hi Bunuel,

What exactly is the solution for:

x^2 x(x-1)0 => x>1 ..OR.. x>0 , x-1 x 0x (signed changed)

I really appreciate your help.

-K
Show more

For any quadratic inequation, here are steps that can be followed to solve it:

1) Write the inequation in the form \(x^2+bx+c>0\) or \(x^2+bx+c0\), we look again at the graph of the parabola and find those values of \(x\) for which the graph is above the Y-axis.
We obtain the solution \(x1.\)

Every upward parabola has its two "arms" above the X-axis for values of \(x\) outside the interval defined by the two roots, and it has the "arc" below the X-axis for values of \(x\) between the two roots.

This is the "justification" for how the sign of a quadratic expression changes in the three intervals defined by the two roots of the quadratic. Once you understand this, it will be really easy to quickly write down the solution for any similar inequation, and you will need just to imagine the parabola.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!