Hi,
So here's my problem:
If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
ax^2 - bx(x-1) + c(x-1)^2 = 0

What I did:
I managed to simplify the given equation to:
x^2(a-b+c) + x(b - 2c) + c = 0

And now I don't know how to proceed.
A slight push in the right direction would help.

Cheers.

2. Originally Posted by Ruler of Hell
Hi,

What I did:
I managed to simplify the given equation to:
x^2(a-b+c) + x(b - 2c) + c = 0
Cheers.
Havent checked you simplification but if it is correct then you problem is simple just use the quadratic formula $\displaystyle ax^2 +bx +c = 0$
you a is (a+b+c)
you b is (b-2c)
Can you take it from here? or more pushing needed? =)

3. Originally Posted by Ruler of Hell
Hi,
So here's my problem:
If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
ax^2 - bx(x-1) + c(x-1)^2 = 0

What I did:
I managed to simplify the given equation to:
x^2(a-b+c) + x(b - 2c) + c = 0

And now I don't know how to proceed.
A slight push in the right direction would help.

Cheers.
If $\displaystyle \alpha$ and $\displaystyle \beta$ are roots of $\displaystyle ax^2+ bx+ c= 0$ then you know that $\displaystyle ax^2+ bx+ c= a(x- \alpha)(x- \beta)= ax^2- a(\alpha+ \beta)x+ a\alpha\beta$.

That tells you that $\displaystyle b= a(\alpha+ \beta)$ and $\displaystyle c= \alpha\beta$.
So what are a- b+ c and b- 2c in terms of a, $\displaystyle \alpha$, and $\displaystyle \beta$?

Once you know that, you should be able to use that to find roots of the second equation in terms of a, $\displaystyle \alpha$, and $\displaystyle \beta$.

4. Originally Posted by Ruler of Hell
Hi,
So here's my problem:
If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
ax^2 - bx(x-1) + c(x-1)^2 = 0

What I did:
I managed to simplify the given equation to:
x^2(a-b+c) + x(b - 2c) + c = 0

And now I don't know how to proceed.
A slight push in the right direction would help.

Cheers.
$\displaystyle (x-\alpha)(x - \beta) = (a-b+c)x^2 + (b-2c)x + c$

This means that

$\displaystyle a-b+c = 1$

$\displaystyle \alpha \beta = c$

$\displaystyle \alpha + \beta = (b-2c)$

5. Originally Posted by Ruler of Hell
Hi,
So here's my problem:
If the equation ax^2 + bx + c = 0 has roots alpha and beta, find the roots of the equation:
ax^2 - bx(x-1) + c(x-1)^2 = 0

What I did:
I managed to simplify the given equation to:
x^2(a-b+c) + x(b - 2c) + c = 0

And now I don't know how to proceed.
A slight push in the right direction would help.

Cheers.
As $\displaystyle a,\ b$ and $\displaystyle c$ are constants, along with $\displaystyle \alpha$ and $\displaystyle \beta$, you can calculate the roots of the second equation using these.

You can also calculate the new roots from the original $\displaystyle \alpha$ and $\displaystyle \beta$ by using...

Sum of new roots $\displaystyle =\frac{b-2c}{a+c-b}$

Product of new roots $\displaystyle =\frac{c}{a+c-b}$

Given that $\displaystyle -(\alpha+\beta)=\frac{b}{a}$ and $\displaystyle \alpha\beta=\frac{c}{a}$

you can write $\displaystyle a$ and $\displaystyle c$ in terms of $\displaystyle b$

then solve for the sum and product of the new roots using $\displaystyle b$.

If you do this, you will find the new roots in terms of the old ones..

the calculations will give

new roots=$\displaystyle \frac{\alpha}{1+\alpha},\ \frac{\beta}{1+\beta}$

6. Thank you Henryt999, HallsofIvy, e^(i*pi) & Archie Meade!

I solved the equation using the methods provided by you using
$\displaystyle a' = (a-b+c)$
$\displaystyle b' = (b-2c)$
$\displaystyle c' = c$

And then used the quadratic equation formula to find the new roots in terms of $\displaystyle {a}, {\alpha}, {\beta}$

The $\displaystyle a$ gets canceled, so we're left with only $\displaystyle {\alpha}, {\beta}$.

The final answer I got is:
$\displaystyle \frac{\alpha}{1+\alpha},\ \frac{\beta}{1+\beta}$

Cheers.