1. ## Easy quadratic solving

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

2. Hi
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 $-\frac{B}{2A}=1$.

3. 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

4. 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 $\frac{c-a}{a-b} = 1 \Rightarrow c-a = a-b \Rightarrow 2a = b+c$

5. 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///

6. 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 : $-\frac{B}{2A}=1$

7. Originally Posted by flyingsquirrel
The polynomial still has two roots : $-\frac{B}{2A}=1$
yes yes I got itbefore only thanks anyywa