+0  
 
+5
139
1
avatar

\(4x^3 + 16x^2 - 9x - 36 >= 0\)Find a solution for 4x^3 + 16x^2 - 9x - 36>/= 0

Guest Mar 14, 2017
Sort: 

1+0 Answers

 #1
avatar+76929 
+5

4x^2 + 16x^2 - 9x - 36 ≥ 0   set this to 0

 

4x^2 + 16x^2 - 9x - 36 = 0

 

Factor

 

4x^2(x + 4) - 9(x + 4) = 0

 

(x + 4) (4x^2 - 9)   = 0

 

(x + 4)(2x + 3) (2x - 3) = 0

 

Set each factor to 0 and solve for x ....and x = -4, x = -3/2  and x = 3/2

 

Look at the graph, here : https://www.desmos.com/calculator/tiouq73sj4

 

Notice that the intervals that make the original inequality true  are [-4, -3/2 ] and [3/2, infinity )

 

 

cool cool cool

CPhill  Mar 14, 2017

8 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details