Quadratic equation having roots are symmetric/non-symmetric function of another quadrating equation

Question

Bit curious to know why these equations tend to behave like this.

Let $f(x)=x^2+3x+2=0•••(1)$ be a function whose two roots are $\alpha=-1$ and $\beta=-2$.

$\therefore \alpha+\beta=-3$ and $\alpha\beta=2$

Another equation whose roots are symmetric function of $\alpha$ and $\beta$,which are $\gamma=\alpha+\frac1\beta=-\frac3 2 $ and $\delta=\beta+\frac1\alpha=-3$ is $2x^2+9x+9=0•••(2)$

To obtain second equation we could simply assign $x$( variable of second equation)$=\gamma=\alpha+\frac1\beta=\frac{\alpha\beta+1} \beta=\frac{2+1} \beta=\frac3\beta =>\beta=\frac3 x$

By putting this value of $\beta$ in the $1^{st}$ equation we ibtain the second equation. Using the value of $\delta$ we will obtain the similar equation.

Upto this I could understand that the function of roots is a link between two equations which is like transformation.

But when the roots aren't symmetric functions then why do we obtain different equations for inserting the functions of the two roots in the first equation? Isn't the function of roots linking both equations?

Answer 1

Another equation whose roots are symmetric function of α and β

But $\alpha+ \dfrac{1}{\beta}$ is not a symmetric function of $\alpha,\beta$.

What is true, however, is that if the new roots can be written as $\gamma=h(\alpha,\beta), \delta=h(\beta,\alpha)$ for some function $h$, then $\gamma+\delta$ and $\gamma\delta$ are symmetric functions of $\alpha,\beta$, so it is possible to construct the equation having $\gamma,\delta$ as roots using the symmetric functions of $\alpha,\beta\,$ i.e. without solving the original equation.

In this example, without using OP's partial substitution trick:

• $\displaystyle \gamma+\delta=\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}=\left(\alpha+\beta\right)\left(1+\frac{1}{\alpha\beta}\right)=(-3)\left(1+\frac{1}{2}\right)=-\frac{9}{2}$

• $\displaystyle \gamma\delta=\alpha\beta+1+1+\frac{1}{\beta\alpha}=2+\alpha\beta+\frac{1}{\alpha\beta}=2+2+\frac{1}{2}=\frac{9}{2}$

It follows that the quadratic with roots $\gamma,\delta$ is $x^2 + \dfrac{9}{2}x+\dfrac{9}{2}=0 \iff 2x^2+9x+9=0\,$.

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[ EDIT ] $\;$ Following up on comments regarding this part of the original post.

To obtain second equation we could simply assign $x$( variable of second equation)$=\gamma=\alpha+\frac1\beta=\frac{\alpha\beta+1} \beta=\frac{2+1} \beta=\frac3\beta =>\beta=\frac3 x$

By putting this value of $\beta$ in the $1^{st}$ equation we ibtain the second equation. Using the value of $\delta$ we will obtain the similar equation.

• The notation would be less confusing if it used a different variable for the transformation, for example $\color{red}{y} = \gamma = \dots \implies \beta=\frac{3}{y} \implies f\left(\frac{3}{y}\right) = 0 \implies 2y^2+9y+9=0$.

• Just the $\gamma=\frac{3}{\beta}$ condition is not enough to derive the equation. For example, it is also true in this case that $\gamma=\frac{3}{2}\alpha$, but substituting $x=\frac{2}{3}y$ in the original equation does not produce the correct result.

  The implicit assumption here is that not only $\gamma = \frac{3}{\beta}$ but also $\delta=\frac{3}{\alpha}$, and this is what allows the substitution to work.

A more direct (and general) way to state it would be: let $x_1=\alpha,x_2=\beta$ be the roots of $f(x)=0$, and let $y_1=\gamma=g(x_1), y_2=\delta=g(x_2)$ for some function $g$. If $g$ is invertible, then the equation having $y_1,y_2$ as roots is $f\left(g^{-1}(x)\right)=0\,$.