The quadratic inequality ax² + bx + 1 > 0 has solution set consisting of all x with −1 < x < 2. Find the value of ab.

Step-by-step solution

1. 1

   Deduce the sign of $a$ from the shape of the solution set. The inequality holds strictly between two numbers. An upward parabola ($a>0$) is negative between its roots and positive outside, so it can never give a bounded solution set for "$>0$". Only a downward parabola can, hence

   $a<0.$

   Getting this sign right is what pins the problem down.

2. 2

   Identify the roots. The solution set is the open interval between the two zeros of $ax^{2}+bx+1$, so those zeros are exactly $x=-1$ and $x=2$. Therefore

   $ax^{2}+bx+1=k(x+1)(x-2)\quad\text{for some }k<0.$

3. 3

   Match the constant term to find $k$. Expanding, $k(x+1)(x-2)=kx^{2}-kx-2k$. The constant terms must agree:

   $-2k=1\;\Longrightarrow\;k=-\frac12,$

   which is indeed negative, consistent with step 1.

4. 4

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

   $ax^{2}+bx+1=-\frac12x^{2}+\frac12x+1,$

   so $a=-\tfrac12$ and $b=\tfrac12$. As a cross-check with Vieta: the product of the roots is $1/a=(-1)(2)=-2$, giving $a=-\tfrac12$, and the sum is $-b/a=-1+2=1$, giving $b=-a=\tfrac12$. Both routes agree.

5. 5

   Compute the product and verify.

   $ab=\left(-\frac12\right)\left(\frac12\right)=-\frac14.$

   Testing the inequality: at $x=0$ (inside the interval) $-\tfrac12(0)+\tfrac12(0)+1=1>0$ ✓; at $x=3$ (outside) $-\tfrac92+\tfrac32+1=-2<0$ ✓; at $x=-1$ and $x=2$ the expression is $0$, so the endpoints are correctly excluded.