If $\alpha,\beta,\gamma,\delta $ are the roots of the equation $x^{4}-2x^{3}+2x^{2}+1=0$, then the equation whose roots are $2+\displaystyle \frac{1}{\alpha},2+\frac{1}{\beta},2+\frac{1}{\gamma},2+\frac{1}{\delta}$, is:
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$x^{4}-8x^{3}+12x^{2}-42x+29=0$
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$x^{4}+8x^3+ 6x^{2}-29=0$
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$x^{4}-14x+29=0$
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$x^{4}-8x^{3}+26x^{2}-42x+29=0$
D
Correct answer
Explanation
Let y = 2 + 1/x, so 1/x = y - 2, or x = 1/(y - 2). Substitute this into the original equation x^4 - 2x^3 + 2x^2 + 1 = 0. Solving this transformation leads to the polynomial x^4 - 8x^3 + 26x^2 - 42x + 29 = 0.