Solving an equation with quadratic and ratio constraints

Question

I'm struggling with a problem that involves finding all real numbers $p$ that satisfy certain conditions related to a quadratic equation. The equation is as follows:

$$p x^2+\left(1+p^2\right) x+p=0$$

The problem asks me to find all values of $p$ for which this equation has two distinct solutions whose ratio is equal to 2. I've attempted to solve this by using the quadratic formula and manipulating the expressions to satisfy the ratio constraint, but I'm not sure if I'm on the right track.

Specifically, how do I use the quadratic formula to find the roots of the equation? I know I have to set the ratio of the two roots equal to 2:

$$\frac{x_1}{x_2}=2$$

Any help would be greatly appreciated. Thanks in advance.

Answer 1

There are 2 cases:
1. $$px^2+(1+p^2)x+p=(px+1)(x+p)=0$$ $$px_1+1=0\quad\text{and}\quad x_2+p=0$$ $$x_1=-\dfrac{1}{p}\quad\text{and}\quad x_2=-p$$ $$\frac{x_2}{x_1}=\dfrac{-p}{-\frac{1}{p}}=p^2=2$$ $$p=\pm\sqrt{2}$$ 2. $$px^2+(1+p^2)x+p=(px+1)(x+p)=0$$ $$px_2+1=0\quad\text{and}\quad x_1+p=0$$ $$x_2=-\dfrac{1}{p}\quad\text{and}\quad x_1=-p$$ $$\frac{x_2}{x_1}=\dfrac{-\frac{1}{p}}{-p}=\dfrac{1}{p^2}=2$$ $$p=\pm\dfrac{1}{\sqrt{2}}$$ So $p\in\left\{\sqrt{2},-\sqrt{2},\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right\}$

Answer 2

To apply the quadratic formula as you ask about, the discriminant of the given polynomial is $ \ \Delta \ = \ (1 + p^2)^2 \ - \ 4·p^2 \ = \ 1 - 2p^2 + p^4 \ = \ (1 - p^2)^2 \ \ . \ $ If we call the roots of the equation $ \ r \ $ and $ \ 2r \ \ , \ $ their difference is $ \ 2r - r \ = \ r \ = \ \frac{ \sqrt{\Delta} }{p} \ = \ \frac{|1 \ - \ p^2|}{p} \ \ . \ $ Since the sum of the two roots is $ \ 3r \ = \ -\frac{1 \ + \ p^2}{p} \ \ , \ $ we have two cases to solve for $ \ r \ $ and $ \ p \ \ . \ $

Alternatively, for the roots $ \ r \ $ and $ \ 2r \ \ , \ $ we have the factorization $ \ px^2 + (1 + p^2)x + p $ $ = \ p·(x - r)·(x - 2r) \ = \ 0 \ \ . \ $ Comparing the coefficients gives us two equations: $$ p \ \ = \ \ p·2r^2 \ \ , \ \ 1 \ + \ p^2 \ \ = \ \ p·(-3r) \ \ . $$

Since we must have $ \ p \ \neq \ 0 \ \ , \ $ the first equation tells us the the two roots are given by $ \ r \ = \ \pm \frac{1}{\sqrt2} \ \ . \ $ Inserting either of these values into the second equation $ \ p^2 + 3rp + 1 \ = \ 0 \ \ $ leads to four possible values for $ \ p \ \ $ (as also shown by Fabio Caiazzo). These correspond to two "upward-" and two "downward-opening" parabolas.