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Question

If α,β,γ,δ are the roots of the equation x42x3+2x2+1=0, then the equation whose roots are 2+1α,2+1β,2+1γ,2+1δ, is:

A
x48x3+12x242x+29=0
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B
x4+8x3+6x229=0
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C
x414x+29=0
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D
x48x3+26x242x+29=0
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Solution

The correct option is D x48x3+26x242x+29=0
f(x)=x42x3+2x2+1
Let the transformed equation be g(y)=0.
If y is a root of the new equation, then
y=2+1x
So, x=1y2
And hence, f(x)=f(1y2)=0
(1y2)42(1y2)3+2(1y2)2+1=0
Multiplying throughout by (y2)4, we get
g(y)=(y2)4+2(y2)22(y2)+1=0

After expansion and rearrangement, we get
g(y)=y48y3+26y242y+29

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