If - - - - - satisfy -1-2 -1-2 -1-2- -1-2- - - - - - 4- If the equation a-0 x-4 - a-1 x-3 -a-2 x-2 - a-3 x-a-4 -0 has the roots - - - 1-1 - - - - 1-1 - - - 1-1 - - - - 1-1 - -then the value of a-2-a-0 is

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Relation Between Roots and Coefficients for Higher Order Equations

If α, β, γ, δ...

Question

If $α, β, γ, δ \in R$ satisfy $\frac{(α + 1)^{2} + (β + 1)^{2} + (γ + 1)^{2} + (δ + 1)^{2}}{α + β + γ + δ} = 4$ .
If the equation $a_{0} x^{4} + a_{1} x^{3} + a_{2} x^{2} + a_{3} x + a_{4} = 0$ has the roots $(α + \frac{1}{β} - 1), (β + \frac{1}{γ} - 1), (γ + \frac{1}{δ} - 1), (δ + \frac{1}{α} - 1)$ , then the value of $\frac{a_{2}}{a_{0}}$ is

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Solution

$(α - 1)^{2} + (β - 1)^{2} + (γ - 1)^{2} + (δ - 1)^{2} = 0$
$α = β = γ = δ = 1$
So, all the roots are (1+1-1)=1
So, the equation is $(x - 1)^{4} = 0$
$x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$
$\frac{a_{2}}{a_{0}} = \frac{6}{1} = 6$

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If α,β,γ,δ∈R satisfy (α+1)2+(β+1)2+(γ+1)2+(δ+1)2α+β+γ+δ=4. If the equation a0x4+a1x3+a2x2+a3x+a4=0 has the roots (α+1β−1),(β+1γ−1),(γ+1δ−1),(δ+1α−1), then the value of a2a0 is

(α−1)2+(β−1)2+(γ−1)2+(δ−1)2=0 α=β=γ=δ=1 So, all the roots are (1+1-1)=1 So, the equation is (x−1)4=0 x4−4x3+6x2−4x+1=0 a2a0=61=6

$(α - 1)^{2} + (β - 1)^{2} + (γ - 1)^{2} + (δ - 1)^{2} = 0$
$α = β = γ = δ = 1$
So, all the roots are (1+1-1)=1
So, the equation is $(x - 1)^{4} = 0$
$x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$
$\frac{a_{2}}{a_{0}} = \frac{6}{1} = 6$