Question

# lf $$\alpha,\beta,\gamma$$ are the roots of the equation $$x^{3}-x+2=0$$, then the equation whose roots are $$\displaystyle \alpha \beta +\cfrac { 1 }{ \gamma }, \displaystyle\beta\gamma+\frac{1}{\alpha},\gamma\alpha+\frac{1}{\beta}$$, is:

A
2y3+y2+1=0
B
2y3y2+1=0
C
y3+y2+1=0
D
2y3+y21=0

Solution

## The correct option is B $$2y^{3}+y^{2}-1=0$$As $$\alpha ,\beta ,\gamma$$ are roots of $${ x }^{ 3 }-x+2=0$$We get $${ s }_{ 3 }=\alpha \beta \gamma =-2$$  ..(1)Now $$y=\alpha \beta +\cfrac { 1 }{ \gamma } \Rightarrow y=\cfrac { \alpha \beta \gamma +1 }{ \gamma }$$ Using (1)$$y=\cfrac { -2+1 }{ \gamma } \Rightarrow y=-\cfrac { 1 }{ x } \Rightarrow x=-\cfrac { 1 }{ y }$$Replacing $$x\rightarrow -\cfrac { 1 }{ y }$$ in $${ x }^{ 3 }-x+2=0$$ , we get$${ \left( -\cfrac { 1 }{ y } \right) }^{ 3 }-\left( -\cfrac { 1 }{ y } \right) +2=0\Rightarrow -1+{ y }^{ 2 }+2{ y }^{ 3 }=0\\ \Rightarrow 2{ y }^{ 3 }+{ y }^{ 2 }-1=0$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More