If - - Y - - R satisfy -12-12- r -12-12-4-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- Y -1- Y -1-1-1-1- then thevalue of a2-an is

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Standard XIII

Mathematics

Higher Order Equations

If α, β, Y , ...

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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Similar questions

If α, β, γ and δ are the roots of $x^{4}$ + 2 $x^{3}$ + 3 $x^{2}$ + 4x + 5 = 0. Find the equation whose roots are $\frac{1}{α}$ , $\frac{1}{β}$ , $\frac{1}{γ}$ , $\frac{1}{δ}$

Q. If

α, β, γ, δ

are the roots of the equation

x^{4} - 2 x^{3} + 2 x^{2} + 1 = 0

, then the equation whose roots are

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, is:

Q. If

α

and

β

are roots of

x^{2} + p x + 1 = 0

, and

γ

and

δ

are the roots of

x^{2} + q x + 1 = 0

show that

q^{2} - p^{2} - (α - γ) (β - γ) (α + δ) (β + δ) = 0

Q. Given that

α, γ

are the roots of the equation

A x^{2} - 4 x + 1 = 0

and

β, δ

the roots of equation,

B x^{2} - 6 x + 1 = 0

where

α, β, γ, δ,

are in H.P. Then

Q. If

α, β

are the roots of

x^{2} + p x + q = 0

and

γ, δ

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x^{2} + p x - r = 0

, then

\frac{α - γ}{β - γ} \cdot \frac{α - δ}{β - δ}

is equal to

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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 $(x - 1)^{4} = 0$
$x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$
$\frac{a_{2}}{a_{0}} = \frac{6}{1} = 6$