• July 2nd 2008, 03:23 AM
ice_syncer
If the roots of the quadratic equation (a – b)x² + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

well, this one's easy, if the equation is of the form Ax^2 +Bx +C=0 then the roots are -B/2A and -B/2A

so the roots are x= -b+c/(2a-2b) and x = -b+c/(2a-2b)

we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help
• July 2nd 2008, 03:36 AM
flyingsquirrel
Hi
Quote:

Originally Posted by ice_syncer
we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help

Notice that $(a-b)\cdot 1^2+(b-c)\cdot 1+c-a = a-b+b-c+c-a=0$ hence one is a root of the polynomial. That's why one has $-\frac{B}{2A}=1$.
• July 2nd 2008, 03:40 AM
topsquark
Quote:

Originally Posted by ice_syncer
If the roots of the quadratic equation (a – b)x² + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

well, this one's easy, if the equation is of the form Ax^2 +Bx +C=0 then the roots are -B/2A and -B/2A

so the roots are x= -b+c/(2a-2b) and x = -b+c/(2a-2b)

we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help

You could look at the discriminant of the original equaion and set it equal to 0. (That's the condition for two equal roots.)

-Dan
• July 2nd 2008, 03:41 AM
Isomorphism
Quote:

Originally Posted by ice_syncer
If the roots of the quadratic equation (a – b)x² + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c.

well, this one's easy, if the equation is of the form Ax^2 +Bx +C=0 then the roots are -B/2A and -B/2A

so the roots are x= -b+c/(2a-2b) and x = -b+c/(2a-2b)

we equate the roots to 1 then we can prove that 2a = b + c but then, how 1 come from ( I GUESSED TO EQUATE TO 1), well, please help

Method 1: (School Method,Straightforward)

If the roots are equal, then the discriminant is 0.

So

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

$b^2 + c^2 -2bc = 4(ac - a^2 -bc + ab)$

$4a^2 + b^2 + c^2 + 2bc - 4ac - 4ab = 0$

$(2a -b-c)^2 = 0$

$2a -b-c = 0$

$2a = b+c$

Method 2: (Simpler, Elegant)

We see that 1 is a root of the equation since $(a-b)(1)^2 + (b-c)1+(c-a) = 0$

So if the roots are equal, the other root also must be 1. This means the product of the roots is 1. Thus $\frac{c-a}{a-b} = 1 \Rightarrow c-a = a-b \Rightarrow 2a = b+c$
• July 2nd 2008, 03:41 AM
ice_syncer
Quote:

Originally Posted by flyingsquirrel
Hi

Notice that $(a-b)\cdot 1^2+(b-c)\cdot 1+c-a = a-b+b-c+c-a=0$ hence one is a root of the polynomial. That's why one has $-\frac{B}{2A}=1$.

But it is a quadratic equation, so it can only have 2 roots at a time and those are the ones I found///
• July 2nd 2008, 04:07 AM
flyingsquirrel
Quote:

Originally Posted by ice_syncer
But it is a quadratic equation, so it can only have 2 roots at a time and those are the ones I found///

The polynomial still has two roots : $-\frac{B}{2A}=1$ :confused:
• July 2nd 2008, 04:53 AM
ice_syncer
Quote:

Originally Posted by flyingsquirrel
The polynomial still has two roots : $-\frac{B}{2A}=1$ :confused:

yes yes I got itbefore only thanks anyywa