How to solve the following quadratic inequalities and write them in the interval notation?
x2 < 25
(2x-3)(2x+5) > -7
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
(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