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Question

If $$ \alpha \neq \beta$$ but $$\alpha^{2}=5\alpha -3$$ and $$\beta^{2}=5\beta-3$$ then the equation having $$\dfrac{\alpha}{\beta}$$ and $$\dfrac{\beta}{\alpha}$$ as its roots is 


A
3x219x+3=0
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B
3x2+19x3=0
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C
3x219x3=0
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D
x25x+3=0
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Solution

The correct option is B $$3x^{2}-19x+3=0$$
$$\alpha^2=5\alpha-3$$                        ------- ( 1 )
$$\beta^2=5\beta-3$$                    ------- ( 2 )
Subtracting ( 2 ) from ( 1 /0,
$$\Rightarrow$$  $$\alpha^2-\beta^2=5\alpha-3-5\beta+3$$
$$\Rightarrow$$  $$(\alpha+\beta)(\alpha-\beta)=5(\alpha-\beta)$$
$$\Rightarrow$$  $$\alpha+\beta=5$$                       ------- ( 3 )
Now, adding ( 1 ) and ( 2 ?),
$$\Rightarrow$$  $$\alpha^2+\beta^2=5\alpha-3+5\beta-3$$
$$\Rightarrow$$  $$\alpha^2+\beta^2=5(\alpha+\beta)-6$$
$$\Rightarrow$$  $$\alpha^2+\beta^2=5(5)-6$$
$$\Rightarrow$$  $$\alpha^2+\beta^2=25-6$$
$$\therefore$$  $$\alpha^2+\beta^2=19$$                ------ ( 4 )
$$\Rightarrow$$  $$(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$$
$$\Rightarrow$$  $$(5)^2=19+2\alpha\beta$$                  [ Using ( 3 ) and ( 4 ) ]
$$\Rightarrow$$  $$25-19=2\alpha\beta$$
$$\Rightarrow$$  $$2\alpha\beta=6$$
$$\therefore$$   $$\alpha\beta=3$$               ------ ( 5 )
Now,
$$\Rightarrow$$  $$\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)=\dfrac{\alpha^2+\beta^2}{\alpha\beta}$$
                            $$=\dfrac{19}{3}$$                   [ Using ( 4 ) and ( 5 ) ]

$$\Rightarrow$$  $$\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)=\dfrac{19}{3}$$                   ----- (  6 )
$$\Rightarrow$$  And $$\left(\dfrac{\alpha}{\beta}\times \dfrac{\beta}{\alpha}\right)=1$$                 ---- ( 7 )
The required equation,
$$x^2-\left(\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}\right)x+\left(\dfrac{\alpha}{\beta}\times \dfrac{\beta}{\alpha}\right)=0$$

$$\Rightarrow$$  $$x^2-\dfrac{19}{3}x+1=0$$             [ From ( 6 ) and ( 7 )]
$$\therefore$$  $$3x^2-19x+3=0$$

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