$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$.

Question

Given solution: Dividing the first equation by $x^2$ we get

$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$

so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}

The answers so far are proving retrospectively that the roots are indeed those given above.I would like to know how the relation of the roots is derived.

Answer 1

The definition says: $t$ is a root of a polynomial $P$ if and only if $P(t)=0$.

so let $P(x)=cx^2+bx+a$, we have : $$P(\frac{1}{x_1})=c(\frac{1}{x_1^2})+b(\frac{1}{x_1})+a=\frac{c+bx_1+ax_1^2}{(x_1)^2}$$

and as we know $x_1$ is a root of $ax^2+bx+c=0$ hence $ax_1^2+bx_1+c=0$ and finally: $P(\frac{1}{x_1})=0$

as a conclusion $\frac{1}{x_1}$ is a root of $cx^2+bx+a$.

You can do the same for $\frac{1}{x_2}$

Answer 2

Given $ax^2 + bx + c = 0$, $$\begin{align} ax^2 + bx + c& = 0\\ a + b\left(\frac{1}{x}\right) + c\left(\frac{1}{x^2}\right) & = 0 \tag{Divide both sides by $x^2$}\\ cy^2 + by + a & = 0\tag{Let $y = \frac{1}{x}$}\end{align}$$ We'd discover that this polynomial has roots $y_1$ and $y_2$. We know that $y = \frac{1}{x}$. So, without loss of generality, assume $y_1 = \frac{1}{x_1}$. Then $y_2 = \frac{1}{x_2}$. These are thus the roots.

Answer 3

Another way to see it is to look at sum and product of roots. $x_1$ and $x_2$ are roots of $ax^2+bx+c=0$ so $x_1+x_2=-\frac{b}{a}$ and $x_1\cdot x_2=\frac{c}{a}$.

$$\frac{a}{c}=\frac{1}{x_1}\cdot \frac{1}{x_2}$$

$$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=-\frac{b}{c}$$

and this shows that $\frac{1}{x_1}$ and $\frac{1}{x_2}$ are the roots of

$$X^2+\frac{b}{c}X+\frac{a}{c}=0$$

Equivalently

$$cX^2+bX+a=0$$