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Question

# 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

A
1619
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B
1619
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C
1916
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D
1916
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Solution

## The correct option is B −1619α,β,γ,δ are the roots of equation 2x4+4x3−3x2+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+y2y)+1=0⇒2(1+y)4+8y(1+y)3−12y2(1+y)2+24y3(1+y)+16y4=0⇒2(y4+4y3+…)+8y(y3+3y2+…) −12y2(y2+2y+1)+24y3(1+y)+16y4=0 This equation has roots as 12α−1,12β−1,12γ−1,12δ−1 So, 12α−1+12β−1+12γ−1+12δ−1=−Coefficient of y3Coefficient of y4=−(8+24−24+24)2+8−12+24+16=−3238=−1619

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