Complex Numbers And Quadratic Equations question 520
Question: If $ \alpha ,\beta $ are the roots of the equation $ ax^{2}+bx+c=0 $ then the equation whose roots are $ \alpha +\frac{1}{\beta } $ and $ \beta +\frac{1}{\alpha } $ , is [RPET 1991]
Options:
A) $ acx^{2}+(a+c)bx+{{(a+c)}^{2}}=0 $
B) $ abx^{2}+(a+c)bx+{{(a+c)}^{2}}=0 $
C) $ acx^{2}+(a+b)cx+{{(a+c)}^{2}}=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Here $ \alpha +\beta =-\frac{b}{a} $ and $ \alpha \beta =\frac{c}{a} $ If roots are $ \alpha +\frac{1}{\beta },\beta +\frac{1}{\alpha }, $ then sum of roots are $ =( \alpha +\frac{1}{\beta } )+( \beta +\frac{1}{\alpha } )=(\alpha +\beta )+\frac{\alpha +\beta }{\alpha \beta }=-\frac{b}{ac}(a+c) $ and product $ =( \alpha +\frac{1}{\beta } )( \beta +\frac{1}{\alpha } ) $ $ =\alpha \beta +1+1+\frac{1}{\alpha \beta }=2+\frac{c}{a}+\frac{a}{c} $ $ =\frac{2ac+c^{2}+a^{2}}{ac}=\frac{{{(a+c)}^{2}}}{ac} $ Hence required equation is given by $ x^{2}+\frac{b}{ac}(a+c)x+\frac{{{(a+c)}^{2}}}{ac}=0 $
Þ $ acx^{2}+(a+c)bx+{{(a+c)}^{2}}=0 $ . Trick: Let $ a=1 $ , $ b=-3,c=2 $ , then $ \alpha =1, $ $ \beta =2 $
$ \therefore \alpha +\frac{1}{\beta }=\frac{3}{2} $ and $ \beta +\frac{1}{\alpha }=3 $ Therefore, required equation must be $ (x-3)(2x-3)=0 $ i.e. $ 2x^{2}-9x+9=0 $ Here (a) gives this equation on putting $ a=1,b=-3,c=2 $ .
BETA
