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# X^2<x absolute value question

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Manager
Joined: 27 Dec 2011
Posts: 70
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05 Oct 2012, 08:09
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)

-K
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Re: X^2<x absolute value question [#permalink]

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05 Oct 2012, 08:22
kartik222 wrote:
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)

-K

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

Hope it helps.
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Re: X^2<x absolute value question [#permalink]

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05 Oct 2012, 10:22
thanks a lot bunuel!!

-K
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Re: X^2<x absolute value question [#permalink]

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07 Oct 2012, 21:43
kartik222 wrote:
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)

-K

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

http://www.veritasprep.com/blog/2012/06 ... e-factors/
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Re: X^2<x absolute value question [#permalink]

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08 Oct 2012, 20:01
Thanks Karishma. Your article explained the concept clearly
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Re: X^2<x absolute value question [#permalink]

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09 Oct 2012, 07:14
2
KUDOS
kartik222 wrote:
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)

-K

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 $$x^2+bx+c<0$$
2) Find the roots of the quadratic equation $$x^2+bx+c=0.$$
3) Sketch the graph of the function $$y=x^2+bx+c$$, which is an upward parabola, intersecting the X-axis at the values found at step 2) above.
4) Read the solution from the graph.

In the beginning, you should practice the steps and draw the graph. Later, you can just imagine it and automatically write down the solution.

Let's apply the above steps to the inequation $$x^2<x.$$
1) $$x^2-x<0$$
2) $$x^2-x=0$$ or $$x(x-1)=0$$; the two roots are 0 and 1.
3) For the graph, see the attached drawing.
4) From the graph, we can see that $$x^2-x$$ is negative (the graph is below the X-axis) if $$x$$ is between the roots 0 and 1.
Therefore, the solution to the given inequation is $$0<x<1.$$

If instead of $$<$$ there was $$\leq$$, the solution would be $$0\leq{x}\leq{1},$$ meaning we should include the values of $$x$$ for which the expression becomes 0. These are exactly the roots of the quadratic equation we solved at step 2).

If we have to solve the inequation $$x^2-x>0$$, 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 $$x<0$$ OR $$x>1.$$

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.
Attachments

Parabola-signs.jpg [ 13.11 KiB | Viewed 890 times ]

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Re: X^2<x absolute value question   [#permalink] 09 Oct 2012, 07:14
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