Forming equation from given roots ($ -\alpha, -\beta $) where $ \alpha,\beta$ are the two roots of $ \ ax^2+bx+c=0$

Question

If this equation $ \ ax^2+bx+c=0 $ has two roots $ \alpha,\beta$ then form an equation which has the roots $ -\alpha,-\beta $

Solution (given): Here, \begin{align} ax^2+bx+c=0 →\alpha,\beta \end{align} So,\begin{align}\ x=\alpha \end{align}

Now the determined equation has the root $ -\alpha $ which is equal to $ -x$.

Now, replacing $ x$ by $ -x$ we get,

\begin{align} a(-x)^2+b(-x)+c=0 \end{align} \begin{align}\Rightarrow ax^2-bx+c=0 \end{align}

But the question is why we put $ -x$ in the previous mentioned equation? I think, that equation belongs to only the roots $ \alpha ,\beta $ . Thus $ a,b,c $ might have different values in determined equation. What I want to say is different roots should occupy different equations and thus different $ a,b,c $ . So we can't put $ -x $ into this equation $ ax^2+bx+c=0 $

Edit

From Adam Rubinson's comment I went through some calculations(thought process actually).I hope you will justify this.I have tried to solve a different question in the following procedure .

(Same type question)If this equation $ \ ax^2+bx+c=0 $ has two roots $ \alpha,\beta$ then form an equation which has the roots $\frac {1-\alpha}{1+\alpha},\frac {1-\beta}{1+\beta} $

Let's imagine, \begin{align} f(x)=ax^2+bx+c \end{align} \begin{align} f(\alpha)=a(\alpha)^2+b(\alpha)+c=0 \end{align}

Now to determine the equation asked in question, we have to just replace $ x$ by g(x) so that putting $\frac {1-\alpha}{1+\alpha}$ in $ g(x)$ yields $ g(\frac {1-\alpha}{1+\alpha})=\alpha $ and thus it becomes $ f(x)=a(g(x) )^2+b(g(x) )+c=0 $

So, \begin{align} g(\frac {1-\alpha}{1+\alpha})=\alpha \end{align} Let, \begin{align} \frac {1-\alpha}{1+\alpha}=m \end{align} \begin{align} \Rightarrow \alpha=\frac {1-m}{1+m} \end{align} \begin{align} \Rightarrow g(m)=\frac {1-m}{1+m} \end{align} \begin{align} \Rightarrow g(x)=\frac {1-x}{1+x} \end{align}

Finally the determined equation is , \begin{align} a(g(x) )^2+b(g(x) )+c=0 \end{align} \begin{align} \Rightarrow a( \frac {1-x}{1+x} )^2+b(\frac {1-x}{1+x} )+c=0 \end{align}

Answer 1

You proved it yourself. If $ax^2+bx+c = 0$ then by putting $-x$ you got $ax^2-bx+c=0$ it means in your case that either $-\alpha$ or $-\beta$ is a root for $ax^2-bx+c$. So you found an equation that has $-\alpha$ and $-\beta$ as a root.

You can also see it by noticing one polynomial with roots $-\alpha$ and $-\beta$ can be written as

$$ a\left(x+\alpha\right)\left(x+\beta\right) = ax^2 +a\left(\alpha + \beta\right)x + a\alpha \beta $$ But you also know that $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$ by root relation in polynomials applied to your first polynomial $x \mapsto ax^2+bx+c$

Answer 2

If $x=\alpha$ solves the equation $f(x)=0,$ then $x=-\alpha$ solves the equation $f(-x)=0.$

As for the second question, your working is correct. The key idea is this: we are given that $f(\alpha)=0.$ Suppose we find a function $g(x)$ such that $g\left(\frac{1-\alpha}{1+\alpha}\right)=\alpha,$ then $\frac{1-\alpha}{1+\alpha}$ is a root of $fg(x) = 0.$ If we define $g(x)$ to be the inverse of $h(x)=\frac{1-x}{1+x},$ then we are guaranteed that $g\left(\frac{1-\alpha}{1+\alpha}\right)=\alpha,$ and so we are done: our equation is $fg(x)=0.$

Answer 3

The function transformation $ \ f(x) \ \rightarrow \ f(-x) \ $ carries out a reflection of the function curve about the $ \ y-$axis ("horizontal flip"). For a parabola described by $ \ f(x) \ = \ ax^2 + bx + c \ \ , $ with $ \ x-$intercepts at $ \ x = \alpha \ $ and $ \ x = \beta \ \ , $ these are reflected into $ \ x = -\alpha \ $ and $ \ x = -\beta \ \ $ by the transformation $ \ f(-x) \ = \ a(-x)^2 + b(-x) + c \ = \ ax^2 - bx + c \ \ . $

We may see this also from the solutions of $ \ ax^2 - bx + c \ = \ 0 \ $ given by the quadratic formula: $$ x \ \ = \ \ -\left(\frac{-b}{2a} \right) \ \pm \ \frac{\sqrt{(-b)^2 \ - \ 4ac}}{2a} \ \ = \ \ - \left( \ \frac{-b}{2a} \ \pm \ \frac{\sqrt{b^2 \ - \ 4ac}}{2a} \ \right) \ = \ -\alpha \ , \ -\beta \ \ . $$

It will be true in general that a polynomial with a given set of roots can be transformed into one with the negatives of those roots by changing the signs of all the coefficients of odd powers of $ \ x \ \ . $