Question #5a205

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Question #5a205

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Answer 1 · 2018-01-04T12:07:52

$- 1 < x < 1$ $-1 < x < 1$

Explanation:

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

subtract 1 from both sides:

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

Factor the left:

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

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

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

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

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

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

Answer 2 · 2018-01-04T18:59:37

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

Explanation:

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

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 $- 1 < - x < 1$ $-1<-x<1$

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Question #5a205

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