Complex Numbers And Quadratic Equations question 705
Question: The roots of the equation $ abc^{2}x^{2}+3a^{2}cx+b^{2}cx-6a^{2}-ab+2b^{2}=0 $ are
Options:
A) non real
B) rational if a, b, c are rational
C) irrational if a, b, c are rational
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The equation is $ {abc^{2}}x^{2}+( 3a^{2}c+b^{2}c )x-6a^{2}-ab+2b^{2}=0 $ Discriminant $ D= {{( 3a^{2}+b^{2} )}^{2}}c^{2}-4abc^{2}( -6a^{2}-ab+2b^{2} ) $ $ = 9a^{4}c^{2} + b^{4}c^{2} + 6a^{2}b^{2}c^{2} + 24a^{3}bc^{2} $ $ +4a^{2}b^{2}c^{2}-8ab^{3}c^{2} $ $ = 9a^{4}c^{2} +16a^{2}b^{2}c^{2} + b^{4}c^{2} + 24a^{3}bc^{2} $ $ -8ab^{3}c^{2}-6a^{2}b^{2}c^{2} $ $ = {{( 3a^{2}c+4abc -b^{2}c )}^{2}} $ Since, the discriminant is a prefect square, therefore the roots are rational provided a, b, c, are rational.
BETA
