Quadratic Equations

**Ruler of Hell** · January 31st 2010, 11:47 AM

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.

**Henryt999** · January 31st 2010, 11:57 AM

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 $ax^2 +bx +c = 0$
you a is (a+b+c)
you b is (b-2c)
Can you take it from here? or more pushing needed? =)

**HallsofIvy** · January 31st 2010, 11:59 AM

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

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

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

**e^(i*pi)** · January 31st 2010, 11:59 AM

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.

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

This means that

$a-b+c = 1$

$\alpha \beta = c$

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

Alternatively use the quadratic formula

**Archie Meade** · January 31st 2010, 12:23 PM

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

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

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

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

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

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

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

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

the calculations will give

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

**Ruler of Hell** · January 31st 2010, 12:46 PM

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

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

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

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

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

Cheers.

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