Question

How do you solve `25-x^2>=0`?

Answer 1

Solution: `-5 <= x <= 5 or [-5,5]`

Explanation:

`25-x^2 >= 0 or (5+x)(5-x)>=0`

Critical points are

`5+x=0 or x= -5 and 5-x=0 or x= 5`

`f(x)=0` when `x=-5 and x=5`

Sign chart:

When `x< -5` sign of `(5+x)(5-x)` is `(-) * (+) = (-) ; < 0`

When `-5 < x < 5` sign of `(5+x)(5-x)` is `(+) * (+) = (+) ; > 0`

When `x> 5` sign of `(5+x)(5-x)` is `(+) * (-) = (-) ; < 0`

Solution: `-5 <= x <= 5 or [-5,5]` [Ans]

Answer 2

`color(blue)([ -5 ,5]`

Explanation:

`25 -x^2 >= 0`

Factor `LHS`:

`-(5+x)(5-x)>=0`

`(5+x)(5-x)<=0`

For:

`5+x<=0`

`x<=-5`

For:

`5-x<=0`

`x>=5`

Solutions:

`color(blue)([ -5 ,5]`