Expert: Paul Klarreich - 5/30/2008

Question
If a, b, and c form an aritmetic sequence, show that the equation(b-c)x^2+(c-a)x+(a-b)=0 has equal roots.  I tried plugging it into the discriminant but the question got really messy and I think i messed something up.  Can you help.

Answer
Questioner:   Patrick
Category:  Advanced Math
Private:  No
 
Subject:  Sequence
Question:  If a, b, and c form an aritmetic sequence, show that the equation(b-c)x^2+(c-a)x+(a-b)=0 has equal roots AND FIND WHAT THEY ARE?  

>> They didn't ask that?  Boy, do they let you off easy.

I tried plugging it into the discriminant but the question got really messy and I think i messed something up.  Can you help
....................................................
Hi, Patrick,

Remember how terms of a G.S. have a common ratio?  Terms of an A.S. have a common difference, which I will call Z. (I would like to use d, but...)

So if a,b,c are an A.S, then
b - a = c - b = Z, the common difference.

a - b = -Z,  b - c = -Z as well.

and

c - a = twice the c.d. = 2Z.

Here is your equation:

(b-c)x^2+(c-a)x+(a-b)=0

Let's use A,B,C for the quadratic coefficients, because we have a,b,c already.

The discriminant is  B^2 - 4AC, which should be zero for equal roots. (I assume you looked that fact up already.)

That is:

(c-a)^2 - 4(b-c)(a-b)

(2Z)^2 - 4(-Z)(-Z)

4Z^2 - 4Z^2 = 0

That wasn't so bad, was it?

Don't like using the discriminant?  Try this:  Let's solve the equation for x:

(b-c)x^2+(c-a)x+(a-b)=0

(-Z)x^2+(2Z)x+(-Z)=0  << put in the c.d.

-Zx^2 + 2Zx - Z = 0  << remove ()

-x^2 + 2x - 1 = 0   << divide out the Z.

x^2 - 2x + 1 = 0  << switch signs.

(x - 1)^2 = 0   << factor

x = 1 and x = 1.  There you are -- equal roots.