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.

What I did so far:

The only time a quadratic equation has no real solutions is when the discriminant is less than $$0$$. In this case, the discriminant equals $$9a^2-4a^2-8a$$, and that is less than $$0$$. Simplifying, I got $$5a^2-8a<0$$, so $$a(5a-8)<0$$. I have $$2$$ cases: $$a<0$$ $$5a-8>0$$ and $$a>0$$ $$5a-8<0.$$ The first case is impossible because $$a$$ must be positive and negative at the same time. The second case says that $$a>0$$ and $$a<\frac{8}{5}$$, which can be written as $$(0,\frac{8}{5})$$ in interval notation, but that answer wasn't correct.

Could I get a little help? Thanks!

• I don't see an obvious problem with your logic. Do you know the official "correct" answer?
– Dan
Commented Jun 6, 2022 at 23:11
• When $a = 0$, the equation becomes $1=0$ which also has no real solution. Commented Jun 6, 2022 at 23:16

There is one subtle special case you're overlooking.

You assume you're dealing with a quadratic, but what if what you're handling is not a quadratic at all, i.e. is something of lesser degree?

Well, then there are two further cases:

• $$f(x) = (a^2+2a)x^2+(3a)x+1$$ is linear (and for simplicity, is not constant)
• $$f$$ is constant

Can either hold, with having no real roots?

In the former case, no. (Note that assuming $$f$$ is not constant is valid, since otherwise just $$f$$ being linear would subsume the constant case, and a distinction needs to be made.) Any line with non-zero slope will cross the $$x$$-axis.

Can $$f$$ be constant? If $$f$$ is constant, then the coefficients of $$x^2,x$$ must be zero. In that event, we would have $$f(x) = 1$$, which definitely has no real roots.

But the question remains, can this occur, and for which $$a$$? Well, the equations

$$a^2 + 2a = 0 \qquad 3a = 0$$

must be satisfied. It is easy to see that this requires $$a=0$$.

Hence, I would imagine the intended answer is actually $$[0,8/5)$$.

Otherwise, your logic holds up. (You can also play with this Desmos demo to see how the function $$f$$ behaves for varying values of $$a$$.)

• Thanks! That really helped :)
– user1043968
Commented Jun 7, 2022 at 0:02
• Oof. Nice answer. (+1) Commented Jun 10, 2022 at 17:51