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quadratic equations
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Author:  kasheee [ Fri, 25 Nov 2005 00:06:46 UTC ]
Post subject:  quadratic equations
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.
Author:  Matt [ Fri, 25 Nov 2005 00:11:11 UTC ]
Post subject: 
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
Author:  jinydu [ Fri, 25 Nov 2005 00:30:36 UTC ]
Post subject:  Re: quadratic equations
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}.
Author:  Matt [ Fri, 25 Nov 2005 00:48:01 UTC ]
Post subject:  Re: quadratic equations
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).
Author:  Soroban [ Fri, 25 Nov 2005 02:35:05 UTC ]
Post subject:  Re: quadratic equations
Hello, kasheee!

Quote:
ax^2+bx+c=0 and cx^2+bx+a=0 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 1.
Author:  kasheee [ Sat, 17 Dec 2005 20:23:15 UTC ]
Post subject: 
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.
Author:  Matt [ Sat, 17 Dec 2005 21:30:20 UTC ]
Post subject: 
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.
Author:  kasheee [ Sun, 30 Jul 2023 16:56:36 UTC ]
Post subject:  Re: quadratic equations
Looking back at this question I posted twenty years ago, could we also relates x1 to y2, and x2 to y1?

Thanks.
Author:  kasheee [ Mon, 31 Jul 2023 20:13:50 UTC ]
Post subject:  Re: quadratic equations
Not neccessary to reply. I can see my question has been answered.
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