Question

# If the sum of the roots of the equation $$ax^{2} + bx + c = 0$$ is equal to sum of the squares of their reciprocals, then $$bc^{2}, ca^{2}, ab^{2}$$ are in

A
A.P
B
G.P
C
H.P
D
A.G.P

Solution

## The correct option is B $$A.P$$Let $$\alpha$$  and $$\beta$$  are the roots of $$a{ x }^{ 2 }+bx+c=0$$We know $$\alpha+\beta = \dfrac { -b }{ a }$$ $$,\alpha .\beta =\dfrac { c }{ a }$$ Given, $$\alpha +\beta =\dfrac { 1 }{ { \alpha }^{ 2 } } +\dfrac { 1 }{ { \beta }^{ 2 } } \quad$$$$=\dfrac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ { \alpha }^{ 2 }.{ \beta }^{ 2 } }$$$$=\dfrac { { \left( \alpha +\beta \right) }^{ 2 }-2\alpha .\beta }{ { \alpha }^{ 2 }{ \beta }^{ 2 } }$$Put values of $$\alpha$$  and $$\beta$$, we get$$\dfrac { -b }{ a } =\dfrac { { \left( \dfrac { -b }{ a } \right) }^{ 2 }-2\dfrac { c }{ a } }{ { \left( \dfrac { { c } }{ { a } } \right) }^{ 2 } }$$$$=\dfrac { \dfrac { { b }^{ 2 } }{ { a }^{ 2 } } -2a\dfrac { c }{ { a }^{ 2 } } }{ \dfrac { { c }^{ 2 } }{ { a }^{ 2 } } }$$$$=\dfrac { { b }^{ 2 }-2ac }{ { c }^{ 2 } }$$$$\Rightarrow -b{ c }^{ 2 }=a{ b }^{ 2 }-2{ a }^{ 2 }c$$$$\Rightarrow 2{ a }^{ 2 }c=a{ b }^{ 2 }+b{ c }^{ 2 }$$Therefore, $$a{ b }^{ 2 },{ a }^{ 2 }c,b{ c }^{ 2 }$$ are in $$A.P.$$Option A is correct.Mathematics

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