How to solve the following quadratic inequalities and write them in the interval notation?

x^{2} < 25

(2x-3)(2x+5) > -7

Guest Jun 8, 2019

#1**+3 **

x^2 < 25

x^2 - 25 < 0

Factor

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

Set each factor to 0 and solve

x + 5 = 0 x - 5 = 0

x = - 5 x = 5

We have three possible intervals that solve this...... (-inf, -5) or (-5, 5) or (5, 0)

What happens is that either the two outside intervals solve the original equation or the middle interval does

Test x = 0, [which is a point in the middle interval ] in the original inequality

0^2 < 25 which is true

So....the interval that solves the inequality is (-5, 5).....we don't include the endpoints because of the "<" sign

Here's a graph : https://www.desmos.com/calculator/fr1w7y9y6f

CPhill Jun 8, 2019

#2**+3 **

(2x-3)(2x+5) > -7 simplify

4x^2 - 6x + 10x - 15 > -7

4x^2 + 4x - 15 > -7 add 7 to both sides

4x^2 + 4x - 8 > 0 divide through by 4

x^2 + x - 2 > 0 factor

(x + 2) ( x - 1) > 0

As before.....set each factor to 0 and solve

x + 2 = 0 x - 1 = 0

x = -2 x = 1

We have three intervals ..... ( -inf, -2) or (-2, 1) or (1, inf)

Test x = 0 in the original inequality and we get that

(2(0)-3)(2(0)+5) > -7

(-3) (5) > -7

-15 > -7

This is not true

So....the intervals that solve this are (-inf, -2) and ( 1, inf)

Here's the graph : https://www.desmos.com/calculator/7dgbgkrmaw

CPhill Jun 8, 2019