# Thread: Can anybody find the Solution Set of this?

1. ## 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

2. draw the graph of $\displaystyle y_1=|x+2|$ and $\displaystyle y_2=|x(x-1)|$ and find for what range of x does $\displaystyle y_1<y_2$. I think this method is much easier.

3. $\displaystyle |x+2|<|x(x-1)|$

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

but $\displaystyle x+2$ is positive if $\displaystyle x>-2$

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

Then both are positive also if $\displaystyle x>1$

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

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

However the right is negative for $\displaystyle 0<x<1$

so then you can write $\displaystyle x+2<-x(x-1)$

4. Here is a third way.
Solve $\displaystyle (x+2)^2<x^2(x-1)^2$

5. 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).

6. 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. ..

7. Yes jam2011,

good method, a variation of Plato's.

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

$\displaystyle |x+2|<|x(x-1)|$

$\displaystyle |x+2|-|x(x-1)|<0$

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

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

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

$\displaystyle 2+x^2>0$

therefore

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

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.

8. Originally Posted by Archie Meade
Yes jam2011,

good method, a variation of Plato's.

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

$\displaystyle |x+2|<|x(x-1)|$

$\displaystyle |x+2|-|x(x-1)|<0$

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

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

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

$\displaystyle 2+x^2>0$

therefore

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

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...