# Comparing coefficients in quadratic?

• October 31st 2013, 08:17 PM
AaPa
Comparing coefficients in quadratic?
Say I have 2 quadratics

p(x^2)+qx=r=0 and d(x^2)+ex+f=0 with one common root

can i write (p/d)=(q/e)=(r/f) ?
• October 31st 2013, 08:31 PM
topsquark
Re: Comparing coefficients in quadratic?
Quote:

Originally Posted by AaPa
Say I have 2 quadratics

p(x^2)+qx=r=0 and d(x^2)+ex+f=0 with one common root

can i write (p/d)=(q/e)=(r/f) ?

Can you?

Compare: $2x^2 + x - 1$ vs. $2x^2 + 3x - 2$

-Dan
• October 31st 2013, 10:43 PM
AaPa
Re: Comparing coefficients in quadratic?
no.
so when can we compare coefficients?
• October 31st 2013, 10:46 PM
SlipEternal
Re: Comparing coefficients in quadratic?
If $p(x),q(x)$ are two polynomials with the exact same set of roots and the same degree, then $p(x) = kq(x)$ where $k$ is some constant. That is when you can compare coefficients.
• November 1st 2013, 02:18 AM
AaPa
Re: Comparing coefficients in quadratic?
All my confusions which troubled me have disappeared. You really made me think on my own.

Thank you very very muchh!!
• November 1st 2013, 02:24 AM
AaPa
Re: Comparing coefficients in quadratic?
hey

like comparing coefficients.

can we compare exponents as in this given equation

[(a^n) + (b^n)]/[(a^n-1) + (b^n-1)] = (a+b)/2
(n-1 is exponent)

to get n=1

if yes then what is the condition that allows us to compare?
• November 1st 2013, 10:03 PM
AaPa
Re: Comparing coefficients in quadratic?
$a^n + b^n / a^n-1 +b^n-1 = a+b / 2$
• November 2nd 2013, 12:06 AM
Idea
Re: Comparing coefficients in quadratic?
Maybe this will help

$x^2 + b x + c$

$x^2 + s x + t$

have a root in common if and only if

$c^2 - b c s + c s^2 + b^2 t - 2 c t - b s t + t^2 = 0$
• November 2nd 2013, 04:26 AM
AaPa
Re: Comparing coefficients in quadratic?
$\frac {a^n + b^n} { a^{n-1} + b^{n-1}} = \frac{a+b} {2}$

by comparing exponents we get n=1
exactly when can we compare exponents like this?
• November 2nd 2013, 05:42 AM
topsquark
Re: Comparing coefficients in quadratic?
Quote:

Originally Posted by AaPa
$\frac {a^n + b^n} { a^{n-1} + b^{n-1}} = \frac{a+b} {2}$

by comparing exponents we get n=1
exactly when can we compare exponents like this?

Well...We can solve this in terms of a and b: b = -a for odd n, so there is no specific relationship needed for n. But generally we can match coefficients of terms on each side of an equation if the terms are members of an orthogonal series. The "power series" is such a series: $a_0 + a_1x + a_2 x^2 + ...$. Many other orthogonal expansions exist but you'll see the power series the most often.

-Dan
• November 2nd 2013, 08:42 AM
AaPa
Re: Comparing coefficients in quadratic?
let us say that a and b are positive distinct real natural numbers....is there any other value of n that satisfies the equation? how do we tell?