We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
44
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+101871 
+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

 

cool cool cool

 Jun 8, 2019
 #2
avatar+101871 
+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

 

 

cool cool cool

 Jun 8, 2019

14 Online Users

avatar