Question

If the root of $x^3-42x^2+336x-512=0$, are in increasing geometric progression, then its common ratio is

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

4

Let the general equation be considered as : $A{x}^3+B{x}^2+Cx+D=0$

Let the roots be $\frac{a}{r},a,ar$. where 'r' is the common ratio.

Now the product of all the roots is
$=\frac{-D}{A}$
$=\frac{-(-512)}{1}$ ...(in the above case)
$=512$.

Hence
$\left(a{r}^{-1}\right)\left(a\right)\left(ar\right)=512$
$a^3=512$
$a=8$
Now sum of all the roots is $=\frac{-B}{A}$
$=42$ ... (in the above case).

Hence
$a+\frac{a}{r}+ar=42$
$\Rightarrow 8+\frac{8}{r}+8r=42$
$\Rightarrow 8r+8+8r^2=42r$
$\Rightarrow 8r^2-34r+8=0$
$\Rightarrow 4r^2-17r+4=0$

$\Rightarrow 4r^2-16r-r+4=0$

$\Rightarrow 4r(r-4)-1(r-4)=0$

$\Rightarrow (4r-1)(r-4)=0$
$r=\frac{1}{4}$ or $r=4$
Hence the roots are
$2,8,32$
Thus the common ratio is 4.