# Thread: Roots of polynomial equations: Substitution method.

1. ## Roots of polynomial equations: Substitution method.

How would you do this one with the substitution method?
Equation
$\displaystyle x^2 + 4x + 7$
Find an equation with roots $\displaystyle \alpha$ + $\displaystyle 2\beta$ and $\displaystyle \beta$ + $\displaystyle 2\alpha$

Example of what i mean:
$\displaystyle x^2 + 5x + 7 = 0$ has roots $\displaystyle \alpha$ and $\displaystyle \beta$
Find an equation with $\displaystyle 2\alpha$ and $\displaystyle 2\beta$
Substitution would be:
Let u = $\displaystyle 2\alpha$
hence $\displaystyle \alpha$ = $\displaystyle \frac{u}{2}$

Sub it into equation gives.
$\displaystyle (\frac{u}{2})^2 + 5(\frac{u}{2}) + 7$
When multiplied out gives
$\displaystyle u^2 + 10u + 28$

Uh, i don't mean to sound patronising when i gave an example since i know a lot of you are muuuch better at maths then me . Just wanted to make sure my question was clear, thats all .

Thankyou very much

2. Originally Posted by AshleyT
Given that $\displaystyle \alpha$ and $\displaystyle \beta$ are roots of the equation $\displaystyle x^2 + 4x + 7 = 0$. Find an equation with roots $\displaystyle \alpha+2\beta$ and $\displaystyle \beta + 2\alpha$
Hello Ashley, A lengthy example is not required you just need to clearly define your question.

Given the root of your quadratic you can write. $\displaystyle x^2 + 4x + 7 = (x -\alpha)(x-\beta)$

Expand and compare coefficients to give.
$\displaystyle \alpha+\beta = -4 \ \ \ (1)$
$\displaystyle \alpha \beta = 7 \ \ \ \ \ \ \ \ (2)$

now you require a quadratic such that $\displaystyle \alpha+2\beta$ and $\displaystyle \beta + 2\alpha$ are roots. So your required quadratic can be written as $\displaystyle (x -( \alpha+2\beta))(x-(\beta + 2\alpha))$

$\displaystyle \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2 \beta^2 + 4 \alpha \beta + 2 \alpha^2$

$\displaystyle \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2(\beta + \alpha)^2$

Bobak

3. Originally Posted by bobak
Hello Ashley, A lengthy example is not required you just need to clearly define your question.

Given the root of your quadratic you can write. $\displaystyle x^2 + 4x + 7 = (x -\alpha)(x-\beta)$

Expand and compare coefficients to give.
$\displaystyle \alpha+\beta = -4 \ \ \ (1)$
$\displaystyle \alpha \beta = 7 \ \ \ \ \ \ \ \ (2)$

now you require a quadratic such that $\displaystyle \alpha+2\beta$ and $\displaystyle \beta + 2\alpha$ are roots. So your required quadratic can be written as $\displaystyle (x -( \alpha+2\beta))(x-(\beta + 2\alpha))$

$\displaystyle \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2 \beta^2 + 4 \alpha \beta + 2 \alpha^2$

$\displaystyle \Rightarrow x^2 -3(\alpha+\beta)x + \alpha \beta +2(\beta + \alpha)^2$

Bobak
Thankyou very much .

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