Problem Solving

Hi,

Can anyone help me out by solving the following problem

(1-x^2) >= 0

(1-x^2) >= 0

this can be simplified to x^2<=1

x lies btw -1 to +1 ( -1<= x <= 1 )

Your equation can be easily written as following:
x^2 <= 1 or x>=-1 and x<=1, hence x belongs to the segment[-1;1]

Thanks Javoni and Shalini for your response.

I had also solved the problem the way u did but I am bit confused with the explanation given in OG 12(this is a problem from OG12). I was expecting some one would solve it that way.

I am mentioning the answer explanation below from OG12 and also underlining the text in which I am confused. Please let me know how it is being derived

"The expression 1-x^2 can be factored as (1-x)(1+x). The product is positive when both the factors are positive (this happens if 1>=x and x>=-1, or equivalently if -1<=x<=1) or both factors are negative (this happens if 1<=x and x <=1, which cannot happen), and therefore the solution is -1<=x<=1"

When both the factors are +ve then the issue will be simplified to -1<=x<=1 but when both the factors are -ve how can it be simplified to 1<=x and x <=1?

"The expression 1-x^2 can be factored as (1-x)(1+x). The product is positive when both the factors are positive (this happens if 1>=x and x>=-1, or equivalently if -1<=x<=1) or both factors are negative (this happens if 1<=x and x <=1, which cannot happen), and therefore the solution is -1<=x<=1"

When both the factors are +ve then the issue will be simplified to -1<=x<=1 but when both the factors are -ve how can it be simplified to 1<=x and x <=1?

When both factors are negative this implies
A) -(1-x)>= 0 and B) -(1+x)>=0

A becomes 1>= x and B becomes x<=-1
But for these values of x the inequality 1-x^2>=0 does not hold good.

SO we consider both (1-x) and (1+x) to be positive.

Thanks Shalini for the explanation..