Question

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 $2+\frac{1}{\alpha },2+\frac{1}{\beta },2+\frac{1}{\gamma },2+\frac{1}{\delta }$, is:

• A. $x^4-8x^3+12x^2-42x+29=0$
• B. $x^4+8x^3+6x^2-29=0$
• C. $x^4-14x+29=0$
• D. $x^4-8x^3+26x^2-42x+29=0$

Answer 1

The correct option is D $x^4-8x^3+26x^2-42x+29=0$

$f\left(x\right)=x^4-2x^3+2x^2+1$
Let the transformed equation be $g\left(y\right)=0$.
If $y$ is a root of the new equation, then
$y=2+\frac{1}{x}$
So, $x=\frac{1}{y-2}$
And hence, $f\left(x\right)=f(\frac{1}{y-2})=0$
$\Rightarrow {(\frac{1}{y-2})}^4-2{(\frac{1}{y-2})}^3+2{(\frac{1}{y-2})}^2+1=0$
Multiplying throughout by $(y-2{)}^4$, we get
$g\left(y\right)=(y-2{)}^4+2(y-2{)}^2-2(y-2)+1=0$

After expansion and rearrangement, we get
$g\left(y\right)=y^4-8y^3+26y^2-42y+29$