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.
+128474 (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
