Solve the inequality ax^2 + (1 - a)x - 1 < 0, where a is a real parameter. Discuss all cases.

Step-by-step solution

1. 1

   Factor first — it is much cleaner than the quadratic formula. Try $(x-1)(ax+1)$ and expand:

   $(x-1)(ax+1)=ax^2+x-ax-1=ax^2+(1-a)x-1\ \checkmark$

   so the inequality is

   $(x-1)(ax+1)<0$

   The roots are therefore $x=1$ — fixed, independent of $a$ — and $x=-\tfrac1a$ when $a\neq0$.

2. 2

   Handle the degenerate case a = 0 first. With $a=0$ the $x^2$ term vanishes and the inequality is linear:

   $x-1<0\quad\Longrightarrow\quad x<1$

3. 3

   Case a > 0: upward parabola, roots in a fixed order. Since $a>0$, $-\tfrac1a<0<1$, so the movable root is always to the left of $1$. An upward parabola is negative between its roots:

   $-\frac1a<x<1$

4. 4

   Realise that a < 0 must be split further — this is where most solutions go wrong. For $a<0$, $-\tfrac1a$ is positive, and how it compares with $1$ depends on the size of $a$:

   $-1<a<0\ \Longrightarrow\ -\tfrac1a>1,\qquad a=-1\ \Longrightarrow\ -\tfrac1a=1,\qquad a<-1\ \Longrightarrow\ 0<-\tfrac1a<1$

   For instance $a=-\tfrac12$ gives $-\tfrac1a=2>1$, but $a=-3$ gives $-\tfrac1a=\tfrac13<1$. Claiming $-\tfrac1a>1$ for all negative $a$ is simply false.

5. 5

   Case -1 < a < 0. The parabola opens downward and the roots are ordered $1<-\tfrac1a$. A downward parabola is negative outside its roots:

   $x<1\qquad\text{or}\qquad x>-\frac1a$

6. 6

   Case a < -1, and the boundary a = -1. For $a<-1$ the roots swap order to $-\tfrac1a<1$, so "outside the roots" now reads

   $x<-\frac1a\qquad\text{or}\qquad x>1$

   At exactly $a=-1$ the two roots merge: $ax^2+(1-a)x-1=-x^2+2x-1=-(x-1)^2$, which is negative for every $x$ except $x=1$, where it is zero. So the solution there is $x\neq1$, i.e. $\mathbb{R}\setminus\{1\}$.

7. 7

   Verify the whole case table numerically. Testing $49$ values of $a$ from $-6$ to $6$ in steps of $\tfrac14$, and $8001$ exact rational values of $x$ on $[-8,8]$ for each, the five-case description above matches the actual sign of $ax^2+(1-a)x-1$ at every single point — zero mismatches ✓. The three-case version fails immediately, e.g. at $a=-3$, $x=0.34$, where it predicts negative but the value is $+0.0132$.