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modulus inequalities

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modulus inequalities [#permalink]  19 Nov 2010, 08:48
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Hi Guys,

x^2 - |x| - 12
--------------- >= 2x
x-3

---------------------------------------------
Consider KUDOS if you like my post!!!
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Re: modulus inequalities [#permalink]  20 Nov 2010, 12:07
4
KUDOS
Expert's post
krishnasty wrote:
Hi Guys,

x^2 - |x| - 12
--------------- >= 2x
x-3

---------------------------------------------
Consider KUDOS if you like my post!!!

I am assuming your question is: \frac{{x^2 - |x| - 12}}{{x - 3}} >= 2x
First of all, remember that we cannot cross multiply in this case. The reason is that we do not know the sign of (x - 3). To handle the 2x on the right hand side, bring it to the left hand side to get:

\frac{{x^2 - |x| - 12}}{{x - 3}} - 2x >= 0

\frac{{-x^2 - |x| + 6x - 12}}{{x - 3}} >= 0

Now take 2 cases.

Case 1: If x >= 0, |x| = x
\frac{{-x^2 - x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 5x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

Now, x^2 - 5x + 12 has no real roots. It will always be positive. (because b^2 - 4ac is negative so we know that it has no real roots. Put x = 0, you get 12 which is positive. So this quadratic will never touch the x axis and will always remain above it.)

So for the left hand side to be negative, x < 3. (x cannot be 3 because in that case the denominator will be 0.)
Since x has to be greater than 0, the only solution here is 0 <= x < 3.

Case 2: If x <= 0, |x| = -x
\frac{{-x^2 + x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 7x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

\frac{{(x - 4)(x - 3)}}{{x - 3}} <= 0
or x <= 4
Since x has to be less than(or equal to) 0, the only solution here is x <= 0.

Combining the solutions above, we get x < 3 as the final solution.

(I have assumed that you are comfortable with quadratic expressions and solving for inequalities once you have factors. If you are not, let me know and I will point you to the related theory.)
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Re: modulus inequalities [#permalink]  20 Nov 2010, 12:13
VeritasPrepKarishma wrote:
krishnasty wrote:
Hi Guys,

x^2 - |x| - 12
--------------- >= 2x
x-3

---------------------------------------------
Consider KUDOS if you like my post!!!

I am assuming your question is: \frac{{x^2 - |x| - 12}}{{x - 3}} >= 2x
First of all, remember that we cannot cross multiply in this case. The reason is that we do not know the sign of (x - 3). To handle the 2x on the right hand side, bring it to the left hand side to get:

\frac{{x^2 - |x| - 12}}{{x - 3}} - 2x >= 0

\frac{{-x^2 - |x| + 6x - 12}}{{x - 3}} >= 0

Now take 2 cases.

Case 1: If x >= 0, |x| = x
\frac{{-x^2 - x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 5x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

Now, x^2 - 5x + 12 has no real roots. It will always be positive. (because b^2 - 4ac is negative so we know that it has no real roots. Put x = 0, you get 12 which is positive. So this quadratic will never touch the x axis and will always remain above it.)

So for the left hand side to be negative, x < 3. (x cannot be 3 because in that case the denominator will be 0.)
Since x has to be greater than 0, the only solution here is 0 <= x < 3.

Case 2: If x <= 0, |x| = -x
\frac{{-x^2 + x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 7x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

\frac{{(x - 4)(x - 3)}}{{x - 3}} <= 0
or x <= 4
Since x has to be less than(or equal to) 0, the only solution here is x <= 0.

Combining the solutions above, we get x < 3 as the final solution.

(I have assumed that you are comfortable with quadratic expressions and solving for inequalities once you have factors. If you are not, let me know and I will point you to the related theory.)

+1 to you .
Thats a Good 1
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Re: modulus inequalities [#permalink]  31 Jan 2011, 00:16
okies, this is a great explanation , concept cleared , thanks karishma. there is one question : when can one be sure that a quadratic expression is always +ve ?
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Re: modulus inequalities [#permalink]  31 Jan 2011, 05:43
Expert's post
garimavyas wrote:
okies, this is a great explanation , concept cleared , thanks karishma. there is one question : when can one be sure that a quadratic expression is always +ve ?

When you see x^2 - 5x + 12, think - will it have roots such that their sum is -5 and product is 12? Since product is +ve, and sum -ve, both roots should be negative. But will any two -ve roots add up to give -5 such that their product is 12? (e.g. -1, -4 or -2, -3 etc we do not get 12 as product) No.

To confirm, notice that b^2 - 4ac (where a = 1, b = -5 and c = 12) is -ve here so it has no real roots. Just put x = 0 and you get x^2 - 5x + 12 = 12 which is positive. This means the graph of this expression lies above the x axis. Remember, it will not meet the x axis anywhere because it has no real roots. Hence x^2 - 5x + 12 will always be positive.
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Re: modulus inequalities [#permalink]  31 Jan 2011, 07:35
hey ,
clearer now, thanks a ton
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Re: modulus inequalities [#permalink]  31 Jan 2011, 08:32
1
KUDOS
Expert's post
garimavyas wrote:
okies, this is a great explanation , concept cleared , thanks karishma. there is one question : when can one be sure that a quadratic expression is always +ve ?

Quadratic expression ax^2+bx+c is always positive when 1. a, the coefficient of x^2, is positive: a>0 AND 2. the equation ax^2+bx+c=0 has no real roots, which happens when discriminant is negative, so when d=b^2-4ac<0.

In this case the graph (parabola) of the function f(x)=ax^2+bx+c is upward and lies above the X-axis, thus is always positive.

Similarly quadratic expression ax^2+bx+c is always negative when 1. a, the coefficient of x^2, is negative: a<0 AND 2. the equation ax^2+bx+c=0 has no real roots, which happens when discriminant is negative, so when d=b^2-4ac<0.

Check Coordinate Geometry chapter of Math Book for more: math-coordinate-geometry-87652.html

Hope it helps.
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Re: modulus inequalities [#permalink]  31 Jan 2011, 08:54
that solidifies the work of Karishma . now it is formulated .

thanks Karishma and Bunuel
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Re: modulus inequalities   [#permalink] 31 Jan 2011, 08:54
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