Question

If the roots of the equation $8x^4-30x^3+35x^2-15x+2=0$ are real and they are in G.P., then the difference between largest and smallest root is

Answer 1

Let the roots be
$\frac{a}{r^3},\frac{a}{r},ar,a{r}^3$
$\frac{a}{r^3}\times \frac{a}{r}\times ar\times a{r}^3=\frac{2}{8}=\frac{1}{4}\Rightarrow a^4=\frac{1}{4}$
$\frac{a}{r^3}\times a{r}^3=\frac{a}{r}\times ar=a^2=\frac{1}{2}$
Quadratic factor corresponding to $\frac{a}{r^3},a{r}^3$ is,
$x^2+px+\frac{1}{2}$
Corresponding to $\frac{a}{r},ar$ is
$x^2+qx+\frac{1}{2}$

$8x^4-30x^3+35x^2-15x+2=8(x^2+px+\frac{1}{2})(x^2+qx+\frac{1}{2})$
$=8x^4+8x^3(p+q)+8x^2(pq+1)+4x(p+q)+2$
On comparing the coefficients of $x^3$ and $x^2$, we get
$p=-\frac{3}{2},q=-\frac{9}{4}$
$\Rightarrow (x^2-\frac{3}{2}x+\frac{1}{2})(x^2-\frac{9}{4}x+\frac{1}{2})=0$
$\Rightarrow (2x^2-3x+1)(4x^2-9x+2)\Rightarrow (2x-1)(x-1)(x-2)(4x-1)=0\Rightarrow x=\frac{1}{4},\frac{1}{2},1,2$
$\therefore \text{Difference}=2-\frac{1}{4}=1.75$

Note: If one can observe $x=1$ as one of the roots, the problem becomes fairly straightforward.