Verified answer

Always solve the equality first: x^2 = 1 => x= -1 or x = 1

Then use common sense: equality with 1 arise at x=1 and -1, and since x^2 is less than 1 between these points, the solution is -1<= x <+ 1

x <= 1^1/2

-1 < x < 1

x≤±1, which is two solutions

x ≤ 1 and x ≤ -1. However, the set of values defined by x ≤ 1 includes those defined by x ≤ -1, so the statement that x ≤ 1 is sufficient to include all solutions

x^2≤1

IxI ≤1

-1≤x≤1

Move the square sign (the ^2) to the other side of the equation. That therefore gives you x is equal to or less than the square ROOT of 1

x^2<=1

x^2-1<=0

(x-1)(x+1)<=0

i.e. x-1<=0; x+1<=0

i.e. x<= 1; x<= -1

the solution set becomes: {infinity, -1]