Complex Numbers And Quadratic Equations question 597
Question: If a and b are the roots of $ 6x^{2}-6x+1=0, $ then the value of $ \frac{1}{2}[ a+b\alpha +c{{\alpha }^{2}}+d{{\alpha }^{3}} ] $ $ \frac{1}{2}[ a+b\alpha +c{{\alpha }^{2}}+d{{\alpha }^{3}} ]+\frac{1}{2}[ a+b\beta +c{{\beta }^{2}}+d{{\beta }^{3}} ] $ is [RPET 2000]
Options:
A) $ \frac{1}{4}(a+b+c+d) $
B) $ \frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4} $
C) $ \frac{a}{2}-\frac{b}{2}+\frac{c}{3}-\frac{d}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \alpha ,\beta $ are the roots of the equation $ 6x^{2}-6x+1=0 $
$ \Rightarrow \alpha +\beta =1,\alpha \beta =1/6 $
$ \therefore \frac{1}{2}[ a+b\alpha +c{{\alpha }^{2}}+d{{\alpha }^{3}} ]+\frac{1}{2}[ a+b\beta +c{{\beta }^{2}}+d{{\beta }^{3}} ] $ = $ a+\frac{1}{2}b(\alpha +\beta )+\frac{1}{2}c({{\alpha }^{2}}+{{\beta }^{2}})+\frac{1}{2}d({{\alpha }^{3}}+{{\beta }^{3}}) $ = $ a+\frac{1}{2}b+\frac{1}{2}c[{{(\alpha +\beta )}^{2}}-2\alpha \beta ]+\frac{1}{2}d[{{(\alpha +\beta )}^{3}} $ $ -3\alpha \beta (\alpha +\beta )] $ = $ a+\frac{b}{2}+\frac{1}{2}c[ {{(1)}^{2}}-2.\frac{1}{6} ]+\frac{1}{2}d[ {{(1)}^{3}}-3.\frac{1}{6} ] $ = $ \frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4} $ .
BETA
