Question

lf $\alpha$ and $\beta$ are the roots of $6x^2-6x+1=0$, then $\frac{1}{2}(a+b\alpha +c{\alpha }^2+d{\alpha }^3)+\frac{1}{2}(a+b\beta +c{\beta }^2+d{\beta }^3)=$

• A. $a+b+c+d$
• B. $a+2b+3c+4d$
• C. $a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}$
• D. $0$

Answer 1

Answer: (C)

$a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}$

If $\alpha$ and $\beta$ are the roots of $6x^2-6x+1$, then
$\alpha +\beta =-\frac{-6}{6}=1$
$\alpha \beta =\frac{1}{6}$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =1-\frac{1}{3}=\frac{2}{3}$
${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=1-3\times \frac{1}{6}=\frac{1}{2}$

Now, $\frac{1}{2}(a+b\alpha +c{\alpha }^2+d{\alpha }^3)+\frac{1}{2}(a+b\beta +c{\beta }^2+d{\beta }^3)$

$=\frac{1}{2}(a+a)+\frac{b}{2}(\alpha +\beta )+\frac{c}{2}({\alpha }^2+{\beta }^2)+\frac{d}{2}({\alpha }^3+{\beta }^3)$

$=a+\frac{b}{2}\times 1+\frac{c}{2}\times \frac{2}{3}+\frac{d}{2}\times \frac{1}{2}$

$=a+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}$