The solution set of the quadratic inequality ax^2 + bx + 1 > 0 is the interval -1 < x < 2. Find the value of ab.

Step-by-step solution

1. 1

   Use the shape of the solution set to fix the sign of $a$. A parabola is positive between its roots only when it opens downward. Here the solution set is the bounded interval $(-1,2)$, i.e. the expression is positive strictly between two numbers, so

   $a<0,$

   and $x=-1$ and $x=2$ must be the two roots of $ax^{2}+bx+1=0$ (the endpoints, where the strict inequality just fails).

2. 2

   Rebuild the quadratic from its roots. Any quadratic with roots $-1$ and $2$ is a multiple of $(x+1)(x-2)$, so

   $ax^{2}+bx+1=k(x+1)(x-2)=k\left(x^{2}-x-2\right)=kx^{2}-kx-2k$

   for some $k<0$ (the sign of $k$ is the sign of $a$).

3. 3

   Pin down $k$ using the constant term. This is the step that makes the problem solvable: the given quadratic has constant term exactly $1$, which is what removes the free scale factor. Comparing constant terms,

   $-2k=1\quad\Longrightarrow\quad k=-\frac{1}{2},$

   which indeed satisfies $k<0$, consistent with step 1.

4. 4

   Read off $a$ and $b$. Substituting $k=-\tfrac12$:

   $ax^{2}+bx+1=-\frac{1}{2}x^{2}+\frac{1}{2}x+1\quad\Longrightarrow\quad a=-\frac{1}{2},\ \ b=\frac{1}{2}.$

5. 5

   Compute $ab$ and verify the original inequality.

   $ab=\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)=-\frac{1}{4}.$

   Check: $-\tfrac12x^{2}+\tfrac12x+1>0$ is equivalent (multiply by $-2$, flipping the sign) to $x^{2}-x-2<0$, i.e. $(x+1)(x-2)<0$, whose solution is exactly $-1<x<2$. Spot values confirm it: at $x=0$ the expression is $1>0$, at $x=\pm$ the endpoints it is $0$, and at $x=3$ it is $-2<0$.