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Math Help - Can anybody find the Solution Set of this?

  1. #1
    rcs
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    Can anybody find the Solution Set of this?

    find the solution set and draw the solution in the real number line.

    |x+2| < |x (x - 1)|

    all i can think is to transpose the right side of the in inequality like
    |x+2| - |x (x - 1)|< 0 equate by zero

    then i am stuck here... if i get the value of x by quadratic ... i don't kn0w to do this as a simple solving.

    thanks
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  2. #2
    Senior Member BAdhi's Avatar
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    draw the graph of y_1=|x+2| and y_2=|x(x-1)| and find for what range of x does y_1<y_2. I think this method is much easier.
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  3. #3
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    |x+2|<|x(x-1)|

    First, notice that x(x-1) is positive if x is negative,

    but x+2 is positive if x>-2

    so you have a bound -2<x<0 for which you can remove the modulus symbols.

    Then both are positive also if x>1

    Both are never negative for the same x, since x<-2 makes the right side positive.

    But you can write -(x+2)<x(x-1) for x<-2

    However the right is negative for 0<x<1

    so then you can write x+2<-x(x-1)
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  4. #4
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    Here is a third way.
    Solve (x+2)^2<x^2(x-1)^2
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  5. #5
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    And a fourth way! First solve the equation |x+ 2|= |x(x-1)|. If x+ 2 and x(x- 1) have the same sign, then that equation becomes x+ 2= x(x- 1)= x^2- x or x^2- 2x= 2. Completing the square, x^2- 2x+ 1= (x- 1)^2= 3 so x= 1+sqrt(3) or x= 1- sqrt(3). If x+ 2 and x(x- 1) have opposite signs, then the equation becomes -(x+ 2)= x(x- 1) or -x- 2= x^2- x so that x^2= -2 which is never true.

    Now the point is that the inequality can change from ">" to "<" and vice-versa only where we have "=" and that is at x= 1- sqrt(3) and 1+ sqrt(3). -1 < 1- \sqrt(3) and if x= -1 the inequality becomes |-1+ 2|< |(-1)(-1-1)| or "1< 2" which is true. That means that the inequality is true for all x< 1- \sqrt{3}.

    Now check a single value of x between 1- sqrt(3) and 1+ sqrt(3) to determine whether or not all points in that interval satisfy the inequality and then check a single value of x larger than 1+ sqrt(3).
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  6. #6
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    lx+2l<lx(x-1l
    lx(x-1l-lx+2l<0
    lx(x-1l^2-lx+2l^2<0
    (x(x-1))-(x+2)( x(x-1)+ (x+2))<0 by x^2-y^2= (x+y)(x-y)
    (x^2-x-x-2) ( x^2-x+x+2)<0
    (x^2-2x-2)(x^2+2)<0

    maybe you can do in this way. ..
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  7. #7
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    Yes jam2011,

    good method, a variation of Plato's.

    However, you've mistyped it on the 2nd line...


    |x+2|<|x(x-1)|

    |x+2|-|x(x-1)|<0

    (x+2)^2-\left[x(x-1)\right]^2<0

    \left[x+2+x(x-1)\right]\left[x+2-x(x-1)\right]<0

    \left(2+x^2\right)\left(2x+2-x^2\right)<0

    2+x^2>0

    therefore

    2x+2-x^2<0\Rightarrow\ x^2-2x-2>0

    Find the roots and the inequality is true for x to the left of the smaller root
    and to the right of the larger root.
    Last edited by Archie Meade; February 7th 2011 at 04:54 PM.
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    Yes jam2011,

    good method, a variation of Plato's.

    However, you've mistyped it on the 2nd line...


    |x+2|<|x(x-1)|

    |x+2|-|x(x-1)|<0

    (x+2)^2-\left[x(x-1)\right]^2<0

    \left[x+2+x(x-1)\right]\left[x+2-x(x-1)\right]<0

    \left(2+x^2\right)\left(2x+2-x^2\right)<0

    2+x^2>0

    therefore

    2x+2-x^2<0\Rightarrow\ x^2-2x-2>0

    Find the roots and the inequality is true for x to the left of the smaller root
    and to the right of the larger root.
    thanks for the correction sir Archie Meade...
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