Complex Numbers And Quadratic Equations question 606
Question: If $ \alpha \ne \beta $ but $ {{\alpha }^{2}}=5\alpha -3 $ and $ {{\beta }^{2}}=5\beta -3 $ , then the equation whose roots are $ \alpha /\beta $ and $ \beta /\alpha $ is [AIEEE 2002]
Options:
A) $ 3x^{2}-25x+3=0 $
B) $ x^{2}+5x-3=0 $
C) $ x^{2}-5x+3=0 $
D) $ 3x^{2}-19x+3=0 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\alpha }^{2}}-5\alpha +3=0 $ …….(i) $ {{\beta }^{2}}-5\beta +3=0 $ …….(ii) From (i) - (ii),
Þ $ ({{\alpha }^{2}}-{{\beta }^{2}})-5\alpha +5\beta =0 $
Þ $ {{\alpha }^{2}}-{{\beta }^{2}}=5(\alpha -\beta ) $
$ \Rightarrow \alpha +\beta =5 $ From (i) + (ii),
Þ $ ({{\alpha }^{2}}+{{\beta }^{2}})-5(\alpha +\beta )+6=0 $
Þ $ ({{\alpha }^{2}}+{{\beta }^{2}})-5.5+6=0 $
Þ $ {{\alpha }^{2}}+{{\beta }^{2}}=19 $ Then $ {{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $
$ \Rightarrow 25=19+2\alpha \beta $
$ \Rightarrow \alpha \beta =3 $ then the equation, whose roots are $ \frac{\alpha }{\beta } $ and $ \frac{\beta }{\alpha } $ , is $ x^{2}-x( \frac{\alpha }{\beta }+\frac{\beta }{\alpha } )+\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=0 $
Þ $ x^{2}-x( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } )+1=0 $
$ \Rightarrow x^{2}-x.\frac{19}{3}+1=0 $
Þ $ 3x^{2}-19x+3=0 $ .
BETA
