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

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 )]$$\therefore$$  $$3x^2-19x+3=0$$Maths

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