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α and β are the roots of x^2+px+1=0 find the equation with the roots α+λand β+λ if λ is a constant,?

and if γ and δ are the roots of x^2+qx+1=0prove that ( α+γ)(β+γ)(α-8)(β-8) = q^2-p^2..

please help ... thanks..

I think the second one would be "+qx" right? : )

First, by Vieta's formulas we have

α + β = -p, αβ = 1

The equation with the roots α + λ and β + λ:

(α + λ) + ( β + λ ) = α + β +2λ = -p + 2λ ,

(α + λ)( β + λ ) = αβ + λ(α + β) + λ² = 1 - pλ + λ²

Thus the equation will be:

x² + (p - 2λ)x + (λ² - pλ + 1) = 0

Similarly we have:

γ + δ = -q, γδ = 1

(α + γ)(β + γ)(α - δ)(β - δ)

= [αβ + γ(α + β) + γ²][αβ - δ(α + β) + δ²]

= (1 - γp + γ²)(1 + δp + δ²)

= (1 - γp)(1 + δp) + (1 - γp)δ² + (1 + δp)γ² + (γδ)²

= 1 + δp - γp - γδp² + δ² - δ²γp + γ² + δpγ² +1

= 2 + δp - γp - p² + δ² - δp + γ² + pγ

= 2 - p² + δ² + γ²

= 2 - p² + (δ + γ)² - 2γδ

= 2 - p² + q² -2

= q² - p²

α and β are the roots of x^2 + px + 1 = 0

α+β = -p, and αβ = 1

The equation with the roots α+λ and β+λ is

[x - (α+λ)][x - (β+λ)] = x^2 - (α+β+2λ)x + (α+λ)(β+λ)

α+β+2λ = 2(λ - p)

(α+λ)(β+λ) = αβ + λ(α+β) + λ^2 = λ^2 - pλ + 1 so the equation is

x^2 - (2λ - p)x + λ^2 - pλ + 1

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Possibly a mis-copy, since you have written the same equation

x^2 + px + 1 = 0, without including q

So, have not attempted that part, but you probably use the same idea,

(sum of the roots, product of roots + manipulation)

Just read other post and noted it should have been q.

Good solution already given

Regards - Ian