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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
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B
Always complex
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C
Opposite in sign
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D
Negative
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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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