If - - - - are the roots of equation 2x4-4x3-3x2-3x-1-0- then value of 12-1- 12-1-12-1-12-1 is

Α, β, γ, δ are the roots of equation 2x^4+4x^3 3x^2+3x+1=0 Let y = 12α 1 ⇒α = y+12y Putting it in the equation, we get 2 (1+y2y )^4+4 (1+y2y)^3 3 (1+y2y )+3 (1 ...

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Standard XI

Mathematics

Relation between Roots and Coefficients for Quadratic

If α, β, γ, δ...

Question

If $α, β, γ, δ$ are the roots of equation $2 x^{4} + 4 x^{3} - 3 x^{2} + 3 x + 1 = 0$ , then value of $\frac{1}{2 α - 1} + \frac{1}{2 β - 1} + \frac{1}{2 γ - 1} + \frac{1}{2 δ - 1}$ is

\frac{16}{19}

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- \frac{16}{19}

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\frac{19}{16}

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- \frac{19}{16}

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Solution

The correct option is B $- \frac{16}{19}$
$α, β, γ, δ$ are the roots of equation $2 x^{4} + 4 x^{3} - 3 x^{2} + 3 x + 1 = 0$
Let $y = \frac{1}{2 α - 1}$
$\Rightarrow α = \frac{y + 1}{2 y}$
Putting it in the equation, we get
$2 {(\frac{1 + y}{2 y})}^{4} + 4 {(\frac{1 + y}{2 y})}^{3} - 3 (\frac{1 + y}{2 y}) + 3 (\frac{1 + y}{2 y}) + 1 = 0 \Rightarrow 2 (1 + y)^{4} + 8 y (1 + y)^{3} - 12 y^{2} (1 + y)^{2} + 24 y^{3} (1 + y) + 16 y^{4} = 0 \Rightarrow 2 (y^{4} + 4 y^{3} + \dots) + 8 y (y^{3} + 3 y^{2} + \dots) - 12 y^{2} (y^{2} + 2 y + 1) + 24 y^{3} (1 + y) + 16 y^{4} = 0$

This equation has roots as $\frac{1}{2 α - 1}, \frac{1}{2 β - 1}, \frac{1}{2 γ - 1}, \frac{1}{2 δ - 1}$
So,
$\frac{1}{2 α - 1} + \frac{1}{2 β - 1} + \frac{1}{2 γ - 1} + \frac{1}{2 δ - 1} = - \frac{Coefficient of y^{3}}{Coefficient of y^{4}} = - \frac{(8 + 24 - 24 + 24)}{2 + 8 - 12 + 24 + 16} = - \frac{32}{38} = - \frac{16}{19}$

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If α,β,γ,δ are the roots of equation 2x4+4x3−3x2+3x+1=0, then value of 12α−1+12β−1+12γ−1+12δ−1 is

If $α, β, γ, δ$ are the roots of equation $2 x^{4} + 4 x^{3} - 3 x^{2} + 3 x + 1 = 0$ , then value of $\frac{1}{2 α - 1} + \frac{1}{2 β - 1} + \frac{1}{2 γ - 1} + \frac{1}{2 δ - 1}$ is