+0  
 
0
49
1
avatar

For what values of a does the equation (a^2 + 2a)x^2 + (3a)x+1 = 0 yield no real solutions x? Express your answer in interval notation.

Guest Jan 15, 2018
Sort: 

1+0 Answers

 #1
avatar+82944 
+1

(a^2 + 2a)x^2 + (3a)x+1 = 0

 

This will  have no real solutions  when

 

(3a)^2 -  4 (a^2 + 2a)*1  < 0     simplify

 

9a^2  -  4a^2 - 8a   <  0

 

5a^2  -  8a  <  0

 

a (5a  -  8 )  <  0     (1)

 

Let us solve this

 

a (5a  -  8)  =  0

 

Setting each factor  to 0   and solve for a

 

a  = 0        a  =  8/5 

 

We  have three  possible solution intervals

 

(-inf, 0)  ,  (0, 8/5)  and  (8/5, inf )     

 

Note that the middle interval  is the only one that makes  (1)  true

 

So......the solution  is    0 ≤ a ≤ 8/5 

 

 

cool cool cool

CPhill  Jan 16, 2018

14 Online Users

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