quadratic equations

kasheee · **Joined:** Thu, 16 Sep 2004 10:35:11 UTC **Posts:** 619

ax^2+bx+c=0 and cx^2+bx+a=0 are 2 quadratic equations with their coefficients in reverse order. Find the relationship between the roots of the 2 equations.
Thanks.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9958

For the first quadratic, the roots are:

$$\rule{0.1pt}{0.1pt}\qquad x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} \quad\text{and}\quad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

and likewise, for the second the quadratic, the roots are:

$$\rule{0.1pt}{0.1pt}\qquad y_1=\frac{-b+\sqrt{b^2-4ac}}{2c} \quad\text{and}\quad y_2=\frac{-b-\sqrt{b^2-4ac}}{2c}$

and thus we obtain the relationships:

$$\rule{0.1pt}{0.1pt}\qquad x_1=\frac{cy_1}{a} \quad\text{and}\quad x_2=\frac{cy_2}{a}$

edit: a/c -> c/a

jinydu · **Joined:** Sun, 7 Nov 2004 08:21:37 UTC **Posts:** 2986

kasheee wrote:

ax^2+bx+c=0 and cx^2+bx+a=0 are 2 quadratic equations with their coefficients in reverse order. Find the relationship between the roots of the 2 equations.
Thanks.

To see very quickly why the roots of those quadratics must be inverses of each other, divide both sides of both equations by x, then use the substitution $u=\frac{1}{x}$ on one and only one of the equations (your choice which one).

As I learned on this very board, polynomial equations of degree n (where n is even) with palindromic (first-to-last the same as last-to-first) coefficients can be reduced to polynomial equations of degree $\frac{n}{2}$ using the substitution $u=x+\frac{1}{x}$ .

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9958

jinydu wrote:

the roots of those quadratics must be inverses of each other

In other words, two more relationships you can derive are:

$\rule{0.1pt}{0.1pt}\qquad x_1y_2=1 \quad\text{and}\quad x_2y_1=1$

(using the notation of my first post above).

Soroban · **Posted:** Fri, 25 Nov 2005 02:35:05 UTC

Hello, kasheee!

Quote:

and

are two quadratic equations
. . with their coefficients in reverse order.
Find the relationship between the roots of the two equations.

The roots of $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ have a product of $\frac{c}{a}$

The roots of $x^2 + \frac{b}{c}x + \frac{a}{c} = 0$ have a product of $\frac{a}{c}$

Therefore, the product of all four roots is

.

kasheee · **Joined:** Thu, 16 Sep 2004 10:35:11 UTC **Posts:** 619

Looking back at the answeras i have received in the past I noticed this question I posted.

Isn't a*x1=y1*c? Not (x1=a*y1)/c

Thanks.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9958

kasheee wrote:

Looking back at the answeras i have received in the past I noticed this question I posted.

Isn't a*x1=y1*c? Not (x1=a*y1)/c

Indeed, kashee. I'm surprised nobody pointed that out earlier. Thanks.

kasheee · **Joined:** Thu, 16 Sep 2004 10:35:11 UTC **Posts:** 619

Looking back at this question I posted twenty years ago, could we also relates x1 to y2, and x2 to y1?

Thanks.

kasheee · **Joined:** Thu, 16 Sep 2004 10:35:11 UTC **Posts:** 619

Not neccessary to reply. I can see my question has been answered.

S.O.S. Mathematics CyberBoard

quadratic equations

Who is online