Question

Let the equation $x^4-16x^3+p{x}^2-256x+q=0$ has 4 positive real roots in G.P, then find $(p+q)$

Answer 1

From the given information regarding the equation, we can infer the following :
If the roots of the equation are $a,ar,a{r}^2,a{r}^3$;
$a+ar+a{r}^2+a{r}^3=16$ i.e. $a(1+r+r^2+r^3)=16$
$a^{2}r+a^2{r}^2+a^2{r}^3+a^2{r}^3+a^2{r}^4+a^2{r}^5=p$ i.e. $a^{2}r(1+r+r^2+2r^3+r^4)=p$
$a^3{r}^3+a^3{r}^4+a^3{r}^5+a^3{r}^6=256$ i.e. $a^3{r}^3(1+r+r^2+r^3)=256$
$a^4{r}^6=q$
Dividing third and first equations, we get $a^2{r}^3$ to be $16$, implying $q$ to be $256$.
Also, $a=\frac{4}{r^{\frac{3}{2}}}$