If the solution set of x² + px + q < 0 is (−1/2, 1/3), find the solution set of qx² + px + 1 > 0.

Step-by-step solution

1. 1

   Turn the given solution set into roots. The leading coefficient of $x^{2}+px+q$ is $+1>0$, so the parabola opens upward and is negative exactly between its two roots. The endpoints of the given interval are therefore the roots:

   $x_1=-\frac12,\qquad x_2=\frac13$

2. 2

   Rebuild the quadratic and read off p and q.

   $x^{2}+px+q=\left(x+\frac12\right)\left(x-\frac13\right)=x^{2}+\left(\frac12-\frac13\right)x-\frac16=x^{2}+\frac16x-\frac16$

   Hence

   $p=\frac16,\qquad q=-\frac16$

   (Vieta gives the same thing: $x_1+x_2=-p$ and $x_1x_2=q$.)

3. 3

   Substitute into the second inequality.

   $-\frac16x^{2}+\frac16x+1>0$

4. 4

   Clear the fractions and fix the sign. Multiply by $6$ (positive, so the direction is unchanged):

   $-x^{2}+x+6>0$

   Now multiply by $-1$, which does reverse the inequality:

   $x^{2}-x-6<0$

5. 5

   Factor and solve.

   $(x-3)(x+2)<0$

   An upward parabola is negative between its roots, so

   $-2<x<3$

   $\boxed{(-2,\,3)}$

6. 6

   Check the endpoints and an interior point. At $x=0$: $q(0)+p(0)+1=1>0$ ✓ (inside). At $x=3$: $-\tfrac16(9)+\tfrac16(3)+1=-\tfrac32+\tfrac12+1=0$, so $x=3$ is excluded, as an open endpoint should be. At $x=4$: $-\tfrac{16}{6}+\tfrac46+1=-\tfrac{6}{6}=-1<0$ ✓ (outside).