Question

If the polynomial $f( x )=4 x ^4- ax ^3+ bx ^2- cx +5$ where $a , b , c \in R$ has four posiive real roots say $r_1, r_2, r_3$ and $r_4$, such that $\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}=1$. Find the value of ' $a$ '.

Answer 1

Consider 4 positive terms
$\frac{r_1}{2}, \frac{r_2}{4}, \frac{r_3}{5}, \frac{r_4}{8}$
A.M. $=\frac{1}{4}\left(\frac{r_1}{2}+\frac{r_2}{4}+\frac{r_3}{5}+\frac{r_4}{8}\right)=\frac{1}{4} \times 1=\frac{1}{4}$
$\text { G.M. }=\left(\frac{r_1}{2} \cdot \frac{r_2}{4} \cdot \frac{r_3}{5} \cdot \frac{r_4}{8}\right)^{1 / 4}=\left(\frac{r_1 \cdot r_2 \cdot r_3 \cdot r_4}{2 \cdot 4 \cdot 5 \cdot 8}\right)^{1 / 4}$
but, $ r _1 r _2 r _3 r _4=\frac{5}{4}$
$\therefore \text { G.M. }=\left[\frac{8}{4(2 \cdot 4 \cdot 8 \cdot 8)}\right]^{1 / 4}=\left(\frac{1}{2^8}\right)^{1 / 4}=\frac{1}{4}$
hence A.M. $=$ G.M.
$\Rightarrow $ All numbers are equal
$ \frac{ r _1}{2}=\frac{ r _2}{4}=\frac{ r _3}{5}=\frac{ r _4}{8}= k $
$ r _1=2 k ; r _2=4 k ; r _3=5 k ; r _4=8 k$
$\Rightarrow \prod r _1=(2 \cdot 4 \cdot 5 \cdot 8) k ^4 $
$ \frac{5}{4}=(2 \cdot 4 \cdot 5 \cdot 8) k ^4$
$\therefore k =1 / 4 $
$\text { hence } r _1=\frac{1}{2} ; r _2=1 ; r _3=\frac{5}{4} ; r _4=2 $
$\Rightarrow \sum r _1=\frac{19}{4} $
$\text { but } r _1+ r _2+ r _3+ r _4=\frac{ a }{4}$
$\Rightarrow \frac{19}{4}=\frac{a}{4} \Rightarrow a=19$