What are the solutions to x^2= 4?

[Kinetica]

Homework Statement

-2≤x²≤4

How to find ...<x<...?
Show I take square root of everything? What about the negative -2?

 

[rootX]

Draw a graph:
y = x^2
y =-2
y = -4

You would see -2≤x2≤4 same as x2≤4

 

[HallsofIvy]

Yes, since $x^2$ is never negative, $-2\le x^2\le 4$ is exactly the same as $0\le x^2\le 4$. But also note that both positive and negative x will give a positive square.

You could also attempt this as two separate inequalities: You should immediately see that $-2\le x^2$ is true for all x. What about $x^2\le 4$? I recommend solving inequalities like this by first solving the related equality. What are the two solutions to $x^2= 4$? Those two numbers (lets call them a and b with a< b) divide the set of all real numbers into 3 intervals: x< a, a< x< b, and b< x. In each of those we have either $x^2< 4$ or $x^2> 4$. You could choose one point in each interval to determine which is true for all points in that interval.