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.
+118667 What have you tried?
I'd try using the quadratic formula. The bit under the square must be great or equal to zero or else the answer/s will not be real.
+845 try subbing it in formula b^2-4ac < 0 ( this represents when a quadratic formula has no real solution)
then rearrange it
+129847 We will have no real solutions if the discriminant is < 0
So
(3a)^2 - 4(a^2 + 2a)(1) < 0
9a^2 - 4a^2 - 8a < 0
5a^2 -8a < 0 factor
a ( 5a -8) < 0
This will = 0 whenever a = 0 or a = 8/5
If a > 8 / 5....it will be positive
And if a < 0 it will be positve
So.....the values of a that produce no real solutions are when a is on the interval (0, 8/5)
