Question

Question #5a205

Answer 1

`-1 < x < 1`

Explanation:

Start with: `x^2<1`

subtract 1 from both sides:

`x^2-1<0`

Factor the left:

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

Find the zeros of the left: `x=-1` and `x=1`.

Create a sign chart for these numbers by testing in the factored form: `(x-1)(x+1)`. I tested `x = -100`, `x=0`, and `x=100`

++++ (-1) ----- (1) +++++

from the sign chart we can see that `(x-1)(x+1)<0` on the interval `-1 < x < 1` and that is the solution set.

If you're familiar with the graph of `y=x^2` you can also graph it along with `y=1` and visually see the solution is where the parabola is less than the line, which is the interval `-1 < x < 1`.

Answer 2

Solution set is (-1,1) or,`-1<-x<-1`

Explanation:

It is a quadratic inequality. Consider intervals `(-oo,-1), (-1,0),(0,1)and (1,oo)`. In equality does not hold good in the intervals `(-oo,-1) and (1,oo)`

By assigning any value to x in the intervals (-1,0) and (0,1), the inequality is found to be valid.
At x=0 also the Inequality holds good.

Thus, inequality holds good in the interval (-1,1). The solution set is thus (-1,1) or `-1<-x<1`