Question

If the roots of ${{x}^{3}}-42{{x}^{2}}+336x-512=0$ are in increasing geometric progression, then its common ratio is
(a) 2
(b) 3
(c) 4
(d) 6

Answer 1

Hint: First, we will assume the roots of cubic equation i.e. three roots which are in geometric progression so, we get as $\dfrac{a}{r},a,ar$ . Now, we will find the sum of the roots of the equation as $\dfrac{-B}{A}$ and the product of the roots as $\dfrac{-D}{A}$ . Our equation is in the given form ${{A}^{3}}+B{{x}^{2}}+Cx+D=0$ . Thus, we will get value of ‘a’ from products of roots and on solving sum of the roots we will get a common ratio value i.e. r. Thus, we will get the answer.

Complete step-by-step answer:
Here, we are given the cubic equation and roots of this equation are in geometric progression (GP) , so we can assume that three roots of this cubic equation are $\dfrac{a}{r},a,ar$ . We know that GP is a sequence of numbers where each term after is found by multiplying the previous one by a fixed, non-zero number called the common ratio. So, in our case the common ratio is ‘r’.
Now, we know that for a cubic equation the sum of the roots of the equation is $\dfrac{-B}{A}$ and the product of the roots is $\dfrac{-D}{A}$ .
So, our equation is in form ${{A}^{3}}+B{{x}^{2}}+Cx+D=0$. We can write it as
Sum of the roots i.e. $\dfrac{a}{r}+a+ar=\dfrac{-B}{A}$ ……………(1)
Product of roots i.e. $\dfrac{a}{r}\cdot a\cdot ar=\dfrac{-D}{A}$ ………………(2)
On putting value in equation (2) where D in out equation is $-512$ and A is 1 we get as
$\dfrac{a}{r}\cdot a\cdot ar=\dfrac{-\left( -512 \right)}{1}=512$
We can also write this equation on cancelling r term as
${{a}^{3}}=512$
Now, we know that the cube of digit 8 is 512. So, we get as
${{a}^{3}}={{8}^{3}}\Rightarrow a=8$ ………………………..(3)
We will now put the value of B i.e. $-42$ and A as 1 in equation (1). We get as
$\dfrac{a}{r}+a+ar=\dfrac{-\left( -42 \right)}{1}=42$
Now, on taking LCM of ‘r’ we get as
$a+ar+a{{r}^{2}}=42r$
We will put the value of ‘a’ from equation (3).
$8+8r+8{{r}^{2}}=42r$
On further simplification, we get as
$8+8r-42r+8{{r}^{2}}=0$
On rearranging the terms, we can write it as
$8{{r}^{2}}-34r+8=0$
Now, we will take 2 common from the whole equation, we get as
$4{{r}^{2}}-17r+4=0$
Now, we will split the middle term in such a way that on multiplying we get a number which is formed by $\left( c\cdot a \right)$ i.e. 16 in this case and on adding we get a middle term i.e. $-17$ . So, we write it as
$4{{r}^{2}}-16r-r+4=0$
Now, we will take 4r common from the first 2 terms and $-1$ from remaining terms. Thus, we get as
$4r\left( r-4 \right)-1\left( r-4 \right)=0$
$\left( 4r-1 \right)\left( r-4 \right)=0$
Thus, on solving we get values of r as $4r-1=0\Rightarrow r=\dfrac{1}{4}$ and $r-4=0\Rightarrow r=4$ .
On taking any value of r and finding roots will give the same answer. So, we will take r as 4 here, and a as 8 and substituting values in $\dfrac{a}{r},a,ar$ . so, we get as
$\dfrac{8}{4},8,8\cdot 4=2,8,32$
Thus, we can say that the common ratio is 4 here.
Hence, option (c) is the correct answer.

Note: We can also take roots as $a,ar,a{{r}^{2}}$ which are in GP but when we will find product of roots then we will not be able to get value of ‘a’ easily as we did in solution i.e. ${{a}^{3}}{{r}^{3}}=512$ . Also, in sum of roots it will be difficult to solve such complex equations i.e. $a+a{{r}^{2}}+a{{r}^{3}}=42$ . So, it's better that we take such roots that are easy for calculation and answer can be obtained correctly.