A) non real
B) rational if a, b, c are rational
C) irrational if a, b, c are rational
D) None of these
Correct Answer: B
Solution :
The equation is \[{{\operatorname{abc}}^{2}}{{x}^{2}}+\left( 3{{a}^{2}}c+{{b}^{2}}c \right)x-6{{a}^{2}}-ab+2{{b}^{2}}=0\] Discriminant \[\operatorname{D}= {{\left( 3{{a}^{2}}+{{b}^{2}} \right)}^{2}}{{c}^{2}}-4ab{{c}^{2}}\left( -\,6{{a}^{2}}-ab+2{{b}^{2}} \right)\] \[= \,9{{a}^{4}}{{c}^{2}} + {{b}^{4}}{{c}^{2}} + 6{{a}^{2}}{{b}^{2}}{{c}^{2}} + 24{{a}^{3}}b{{c}^{2}}\]\[+\,4{{a}^{2}}{{b}^{2}}{{c}^{2}}-8a{{b}^{3}}{{c}^{2}}\] \[= 9{{a}^{4}}{{c}^{2}} +16{{a}^{2}}{{b}^{2}}{{c}^{2}} + {{b}^{4}}{{c}^{2}} + 24{{a}^{3}}b{{c}^{2}}\]\[-\,8a{{b}^{3}}{{c}^{2}}-6{{a}^{2}}{{b}^{2}}{{c}^{2}}\]\[= {{\left( 3{{a}^{2}}c+4abc -{{b}^{2}}c \right)}^{2}}\] Since, the discriminant is a prefect square, therefore the roots are rational provided a, b, c, are rational.
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