Complex Numbers And Quadratic Equations question 570
Question: If the roots of $ ax^{2}+bx+c=0 $ are $ \alpha ,\beta $ and the roots of $ Ax^{2}+Bx+C=0 $ are $ \alpha -k,\beta -k, $ then $ \frac{B^{2}-4AC}{b^{2}-4ac} $ is equal to [RPET 1999]
Options:
A) 0
B) 1
C) $ {{( \frac{A}{a} )}^{2}} $
D) $ {{( \frac{a}{A} )}^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{(\alpha -\beta )}^{2}}={{(\alpha +\beta )}^{2}}-4\alpha \beta =(b^{2}-4ac)/a^{2} $ ……(i) Also $ {{{ (\alpha -k)-(\beta -k) }}^{2}} $ = $ {{{(\alpha -k)+(\beta -k)}}^{2}}-4(\alpha -k)(\beta -k) $ = $ {{(-B/A)}^{2}}-4(C/A) $ $ =(B^{2}-4AC)/A^{2} $ …..(ii) From (i) and (ii), $ (b^{2}-4ac)/a^{2}=(B^{2}-4AC)/A^{2} $ \ $ \frac{b^{2}-4AC}{b^{2}-4ac}={{( \frac{A}{a} )}^{2}} $
BETA
