Question

If the polynomial $f(x)=4x^4-a{x}^3+b{x}^2-cx+5$ 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

Answer: 9

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
$\frac{r_1}{2}=\frac{r_2}{4}=\frac{r_3}{5}=\frac{r_4}{8}=k$
$r_1=2k;r_2=4k;r_3=5k;r_4=8k$
$\Rightarrow \prod r_1=(2\cdot 4\cdot 5\cdot 8)k^4$
$\frac{5}{4}=(2\cdot 4\cdot 5\cdot 8)k^4$
$\therefore k=1/4$
$\text{ hence }{r}_1=\frac{1}{2};r_2=1;r_3=\frac{5}{4};r_4=2$
$\Rightarrow \sum r_1=\frac{19}{4}$
$\text{ but }{r}_1+r_2+r_3+r_4=\frac{a}{4}$
$\Rightarrow \frac{19}{4}=\frac{a}{4}\Rightarrow a=19$