Question

# Let $$\alpha, \beta$$ be the roots of the equation $$ax^2+bx+c=0$$ and $$\alpha^4+\beta^4$$ be the roots of the equation $$px^2+qx+r=0$$, then the roots of the equation $$a^2px^2-4acpx+2c^2p+a^2q=0$$ are:

A
Always +ve
B
Always complex
C
Opposite in sign
D
Negative

Solution

## The correct option is C Opposite in sign$$a{ x }^{ 2 }+bx+c=0$$ ......... $$(i)$$$$\alpha , \beta$$ are the roots of above equation$$\therefore \alpha +\beta =\cfrac { -b }{ a }$$ ..... $$(ii)$$$$\alpha \beta =\cfrac { c }{ a }$$ ........ $$(iii)$$$${ \alpha }^{ 4 }$$ and $${ \beta }^{ 4 }$$ are the roots of equation  $$p{ x }^{ 2 }+qx+r=0$$$$\therefore { \alpha }^{ 4 }+{ \beta }^{ 4 }=\cfrac { -q }{ p }$$ .... $$(iv)$$$${ \alpha }^{ 4 }{ \beta }^{ 4 }=\cfrac { r }{ p }$$ ....... $$(v)$$$${ a }^{ 2 }p{ x }^{ 2 }-4acpx+2{ c }^{ 2 }p+{ a }^{ 2 }q=0$$Let $$r$$ and $$\delta$$ be roots of above equation $$\therefore \quad r+\delta =\cfrac { 4acp }{ { a }^{ 2 }p } =\cfrac { 4c }{ a }$$$$r\delta =\cfrac { 2{ c }^{ 2 }p+{ a }^{ 2 }q }{ { a }^{ 2 }p }$$$$=2{ \left( \cfrac { c }{ a } \right) }^{ 2 }+\cfrac { q }{ p }$$$$=2{ \alpha }^{ 2 }{ \beta }^{ 2 }-\left( { \alpha }^{ 4 }+{ \beta }^{ 4 } \right)$$ ..... [From equation $$(iii)$$ and $$(iv)$$]$$=-\left( { \alpha }^{ 4 }+{ \beta }^{ 4 }-2{ \alpha }^{ 2 }{ \beta }^{ 2 } \right)$$$$=-{ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) }^{ 2 }$$$${ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) }^{ 2 }>0$$$$\therefore -{ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) }^{ 2 }<0$$$$\therefore r\delta <0$$$$\Rightarrow \left( r<0\quad \& \quad \delta >0 \right)$$ or $$\left( r>0\quad \& \quad \delta <0 \right)$$$$\therefore$$ Roots of given equation are opposite in sign.Maths

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