# Question #5a205

Jan 4, 2018

$- 1 < x < 1$

#### Explanation:

Start with: ${x}^{2} < 1$

subtract 1 from both sides:

${x}^{2} - 1 < 0$

Factor the left:

$\left(x - 1\right) \left(x + 1\right) < 0$

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

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

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

from the sign chart we can see that $\left(x - 1\right) \left(x + 1\right) < 0$ 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

Solution set is (-1,1) or,$- 1 < - x < - 1$
It is a quadratic inequality. Consider intervals $\left(- \infty , - 1\right) , \left(- 1 , 0\right) , \left(0 , 1\right) \mathmr{and} \left(1 , \infty\right)$. In equality does not hold good in the intervals $\left(- \infty , - 1\right) \mathmr{and} \left(1 , \infty\right)$
Thus, inequality holds good in the interval (-1,1). The solution set is thus (-1,1) or $- 1 < - x < 1$