Solve the inequality a*x^2 + 2x + 1 > 0 for x, treating a as a real parameter, and give every case.

Step-by-step solution

1. 1

   Compute the discriminant once. For $f(x) = ax^2 + 2x + 1$ (when $a \ne 0$),

   $\Delta = 2^2 - 4a(1) = 4 - 4a = 4(1-a)$

   The sign of $\Delta$ and the sign of $a$ together determine everything, so the case split runs on those two quantities.

2. 2

   Handle the degenerate case a = 0 first. The quadratic term vanishes and the inequality becomes linear:

   $2x + 1 > 0 \quad\Longrightarrow\quad x > -\frac{1}{2}$

   Forgetting that $a$ may be zero is the classic omission in parameter problems — the leading coefficient is not guaranteed non-zero.

3. 3

   Case a > 1 (upward parabola, no real roots). Here $\Delta = 4(1-a) < 0$ and $a > 0$, so the parabola opens up and never touches the axis:

   $ax^2 + 2x + 1 > 0 \ \text{ for all } x \in \mathbb{R}$

4. 4

   Case a = 1 (perfect square). $\Delta = 0$ and $x^2 + 2x + 1 = (x+1)^2$, which is positive everywhere except at its single zero:

   $x \ne -1$

5. 5

   Case 0 < a < 1 (upward parabola, two roots). Now $\Delta > 0$ and the roots are

   $x = \frac{-2 \pm \sqrt{4-4a}}{2a} = \frac{-1 \pm \sqrt{1-a}}{a}$

   An upward parabola is positive outside its roots:

   $x < \frac{-1-\sqrt{1-a}}{a} \quad \text{or} \quad x > \frac{-1+\sqrt{1-a}}{a}$

6. 6

   Case a < 0 (downward parabola). Then $1 - a > 1$, so $\sqrt{1-a} > 1$ and $\Delta > 0$: two real roots again. A downward parabola is positive between its roots. Because dividing by the negative $a$ reverses their order, $\tfrac{-1+\sqrt{1-a}}{a} < 0 < \tfrac{-1-\sqrt{1-a}}{a}$, and the solution is

   $\frac{-1+\sqrt{1-a}}{a} < x < \frac{-1-\sqrt{1-a}}{a}$

   Check with $a = -1$: the roots are $\tfrac{-1 \pm \sqrt2}{-1}$, i.e. $-0.414214$ and $2.414214$, and at the midpoint $x = 1$ the expression is $-1 + 2 + 1 = 2 > 0$, confirming the interval.