Question #5a205

2 Answers
Jan 4, 2018

Answer:

#-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#.

Jan 4, 2018

Answer:

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#