Question

help plz

Solve the inequality |z2 - 11+12 > 1.

Answer 1

Let's find the critical points of the inequality.

|x^2−1|+|x|=1

|x^2−1|+|x|+−|x|=1+−|x| (Add -|x| to both sides)

|x^2−1|=−|x|+1

Either x^2−1=−|x|+1 or x^2−1=−(−|x|+1)

Part 1: x^2−1=−|x|+1

(Flip the equation)

−|x|+1=x^2−1

−|x|+1+−1=x^2−1+−1(Add -1 to both sides)

−|x|=x^2−2

−|x|

−1

=

x^2−2

−1

(Divide both sides by -1)

|x|=−x^2+2

We know either x=−x^2+2 orx −(−x^2+2)

x=−x^2+2 (Possibility 1)

x−(−x^2+2)=−x^2+2−(−x^2+2) (Subtract -x^2+2 from both sides)

x^2+x−2=0

(x−1)(x+2)=0(Factor left side of equation)

x−1=0 or x+2=0(Set factors equal to 0)

x=1 or x=−2

x=−(−x^2+2)(Possibility 2)

x=x^2−2(Simplify both sides of the equation)

x−(x^2−2)=x^2−2−(x^2−2)(Subtract x^2-2 from both sides)

−x^2+x+2=0

(−x−1)(x−2)=0(Factor left side of equation)

−x−1=0 or x−2=0(Set factors equal to 0)

x=−1 or x=2

Check answers. (Plug them in to make sure they work.)

x=1(Works in original equation)

x=−2(Doesn't work in original equation)

x=−1(Works in original equation)

x=2(Doesn't work in original equation)

Part 2: x^2−1=−(−|x|+1)

(Flip the equation)

|x|−1=x^2−1

|x|−1+1=x^2−1+1(Add 1 to both sides)

|x|=x^2

We know eitherx=x^2 orx −x^2

x=x^2(Possibility 1)

x−x^2=x^2−x^2(Subtract x^2 from both sides)

−x^2+x=0

x(−x+1)=0(Factor left side of equation)

x=0 or −x+1=0(Set factors equal to 0)

x=0 or x=1

x=−x^2(Possibility 2)

x−(−x^2)=−x^2−(−x^2)(Subtract -x^2 from both sides)

x^2+x=0

x(x+1)=0(Factor left side of equation)

x=0 or x+1=0(Set factors equal to 0)

x=0 or x=−1

Check answers. (Plug them in to make sure they work.)

x=0(Works in original equation)

x=1(Works in original equation)

x=0(Works in original equation)

x=−1(Works in original equation)

Critical points:

x=1 or x=−1 or x=0

Check intervals in between critical points. (Test values in the intervals to see if they work.)

x<−1(Works in original inequality)

−1<x<0(Works in original inequality)

0<x<1(Works in original inequality)

x>1(Works in original inequality)

Answer:

x<−1 or −1<x<0 or 0<x<1 or x>1