# Roots of Absolute Value with inequalities

• Aug 2nd 2012, 06:11 PM
rcs
Roots of Absolute Value with inequalities
I just wonder why when i solved

|x+2|^2 < |x( x - 1)|^2
.
.
.( x +2 )^2 - ( x^2 - x) ^2 < 0
.( 2 + 2x - x^2) ( x^2+2) < 0
.- ( x^2 - 2x - 2) ( x^2 + 2) < 0
since it is < 0 then i have to reject the ( x^2 + 2)
. then i have left (x^2 - 2x - 2) > 0
. square root of (x - 1)^2 > square root of 3
. |x -1| > sq. root 3

then i finally have these roots x > 1 + sq. root of 3 or x > 1 - sq. root of 3

and my classmate said that the other has to be x < 1 - sq. root of 3... i wonder why it should reverse.
that is my problem...
i need guide here.

Thanks
• Aug 2nd 2012, 07:02 PM
Prove It
Re: Roots of Absolute Value with inequalities
Quote:

Originally Posted by rcs
I just wonder why when i solved

|x+2|^2 < |x( x - 1)|^2
.
.
.( x +2 )^2 - ( x^2 - x) ^2 < 0
.( 2 + 2x - x^2) ( x^2+2) < 0
.- ( x^2 - 2x - 2) ( x^2 + 2) < 0
since it is < 0 then i have to reject the ( x^2 + 2)
. then i have left (x^2 - 2x - 2) > 0
. square root of (x - 1)^2 > square root of 3
. |x -1| > sq. root 3

then i finally have these roots x > 1 + sq. root of 3 or x > 1 - sq. root of 3

and my classmate said that the other has to be x < 1 - sq. root of 3... i wonder why it should reverse.
that is my problem...
i need guide here.

Thanks

Once you get to \displaystyle \displaystyle \begin{align*} -\left(x^2 - 2x - 2\right)\left(x^2 + 2\right) &< 0 \end{align*} we have

\displaystyle \displaystyle \begin{align*} \left(x^2 - 2x - 2\right)\left(x^2 + 2\right) &> 0 \textrm{ and since the second term is always positive, we can divide both sides by it and keep the inequality intact...} \\ x^2 - 2x - 2 &> 0 \\ x^2 - 2x + (-1)^2 - (-1)^2 - 2 &> 0 \\ (x - 1)^2 - 3 &> 0 \\ (x - 1)^2 &> 3 \\ |x - 1| &> \sqrt{3} \\ x - 1 < -\sqrt{3} \textrm{ or } x - 1 &> \sqrt{3} \\ x < 1 - \sqrt{3} \textrm{ or } x &> 1 + \sqrt{3} \end{align*}
• Aug 3rd 2012, 05:26 AM
HallsofIvy
Re: Roots of Absolute Value with inequalities
Quote:

Originally Posted by rcs
I just wonder why when i solved

|x+2|^2 < |x( x - 1)|^2
.
.
.( x +2 )^2 - ( x^2 - x) ^2 < 0
.( 2 + 2x - x^2) ( x^2+2) < 0
.- ( x^2 - 2x - 2) ( x^2 + 2) < 0
since it is < 0 then i have to reject the ( x^2 + 2)
. then i have left (x^2 - 2x - 2) > 0
. square root of (x - 1)^2 > square root of 3
. |x -1| > sq. root 3

then i finally have these roots x > 1 + sq. root of 3 or x > 1 - sq. root of 3

If these were correct, you wouldn't need both. "x> 1- sq. root of 3" would be sufficient.

Quote:

and my classmate said that the other has to be x < 1 - sq. root of 3... i wonder why it should reverse.
that is my problem...
i need guide here.

Thanks
Did you check your answer? If x= 0, it satisfies x< 1+ sqrt(3) so should be in the solution set but |0+2|^2= 4 while |0(0-1)|^2= 0. That does NOT satisfy the inequality. Your error is in thinking that [b]nonlinear[b] inequalities can be treated like linear inequalities. If $\displaystyle x^2> a$ it is NOT true that $\displaystyle x>\sqrt{a}$.

The best way to solve more complicated inequalities is to first solve the associated equation. If you solve |x+ 2|^2= |x(x-1)|^2 you get, of course, x= 1+ sqrt(3) and x= 1- sqrt(3). Because the function is continuous, in order to go from "+" to "-", the function value must pass through "0" so those two points separate ">" from "<". Check a single point in each of the three intervals, say, -3< 1- sqrt(3), 1-sqrt(3)< 0< 1+ sqrt(3), and 1+sqrt(3)< 3, to see which satisfy the inequality and which do not.