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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
 #1
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(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

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