4
$\begingroup$

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!

$\endgroup$
2
  • 1
    $\begingroup$ I don't see an obvious problem with your logic. Do you know the official "correct" answer? $\endgroup$
    – Dan
    Commented Jun 6, 2022 at 23:11
  • 4
    $\begingroup$ When $a = 0$, the equation becomes $1=0$ which also has no real solution. $\endgroup$
    – peterwhy
    Commented Jun 6, 2022 at 23:16

1 Answer 1

Reset to default
6
$\begingroup$

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 $a$.)

$\endgroup$
2
  • 1
    $\begingroup$ Thanks! That really helped :) $\endgroup$
    – user1043968
    Commented Jun 7, 2022 at 0:02
  • 1
    $\begingroup$ Oof. Nice answer. (+1) $\endgroup$
    – insipidintegrator
    Commented Jun 10, 2022 at 17:51

You must log in to answer this question.