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 Mar 1, 2019

#1**+2 **

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.

Melody Mar 1, 2019

#2**0 **

try subbing it in formula b^2-4ac < 0 ( this represents when a quadratic formula has no real solution)

then rearrange it

YEEEEEET Mar 1, 2019

#3**+1 **

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)

CPhill Mar 2, 2019