+0  
 
0
132
2
avatar

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

x2 < 25

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

 Jun 8, 2019
 #1
avatar+106533 
+2

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

 

cool cool cool

 Jun 8, 2019
 #2
avatar+106533 
+2

(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

 

 

cool cool cool

 Jun 8, 2019

19 Online Users