If the solution set of the quadratic inequality $ax^2+bx+c>0$ is $(2,3)$, find the solution set for $cx^2+bx+a<0$

Question

From what I could gather

$$a<0$$ $$2<\frac{-b}{2a} <3$$

$$\implies b>0$$

And $$b^2-4a c>0$$ $$\frac{b^2}{4a}<c$$

Also $$f(0)<0$$ $$\implies c<0$$

But I am not able to relate this data to the new equation. How should it be done

Answer 1

If the set where $ax^2+bx+c > 0$ is $(2,3)$, the only possibility is that $$ax^2+bx+c = -(x-2)(x-3)$$

So $$ax^2+bx+c = -x^2 +5x-6$$

So $a=-1$, $b=5$ and $c=-6$.

Therefore you want to solve $-6x^2 +5x-1 < 0$. You can easily see that the roots of the new polynomials are $1/2$ and $1/3$, and therefore the set where $-6x^2 +5x-1 < 0$ is $(-\infty, 1/3) \cup (1/2, +\infty)$.

Answer 2

$ax^2+bx+c>0$ for x in (2,3)

replace x by $\frac{1}{x}$

thus $ c x^2+bx+a>0$ for x in ($\frac{1}{3},\frac{1}{2}$)

or $cx^2+bx+a<0$ for x in $(-\infty, 1/3) \cup (1/2, +\infty)$.

Answer 3

We may also use the "vertex form" for the parabola here. The axis of symmetry is midway between the zeroes of the quadratic polynomial $ \ ax^2 + bx + c \ \ , $ then we can express it as $ \ a·\left(x - \frac52 \right)^2 \ + \ c - \frac{5^2·a}{2^2} \ \ . $ The two zeroes are $ \ \frac12 \ $ unit away from the symmetry axis, so we find that $$ a·\left(\pm \frac12 \right)^2 \ + \ c - \frac{25·a}{4} \ \ = \ \ 0 \ \ \Rightarrow \ \ a \ + \ 4c \ \ = \ \ 25a \ \ \Rightarrow \ \ c \ \ = \ \ 6a \ \ . $$ The "standard form" for the polynomial is thus $ \ ax^2 \ - \ 5ax \ + \ 6a \ \ ; $ in order to recover the given inequality condition, we only need $ \ a < 0 \ \ $ (a "downward-opening" parabola).

Reversing the order of the coefficients produces $$ \ 6ax^2 \ - \ 5ax \ + a \ \ = \ \ a·(6x^2 - 5a + 1) \ \ = \ \ a·(3x - 1)·(2x - 1) \ \ , $$ for which the zeroes are $ \ x \ = \ \frac13 \ , \ \frac12 \ \ . $ (In fact, there is a familiar theorem -- to which I believe Albus Dumbledore is alluding -- that reversing the order of the coefficients of a polynomial produces zeroes which are reciprocals of the original zeroes.) Since we are keeping $ \ a < 0 \ \ , $ the new parabola also "opens downward". As we are now asked for the intervals in which the function is negative, these are $ \ x < \frac13 \ , \ x > \frac12 \ \ . $