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Question: If the roots of \({{x}^{3}}-42{{x}^{2}}+336x-512=0\) are in increasing geometric progression, then i...

If the roots of x3−42x2+336x−512=0{{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

Explanation

Solution

First, we will assume the roots of cubic equation i.e. three roots which are in geometric progression so, we get as ar,a,ar\dfrac{a}{r},a,ar . Now, we will find the sum of the roots of the equation as −BA\dfrac{-B}{A} and the product of the roots as −DA\dfrac{-D}{A} . Our equation is in the given form A3+Bx2+Cx+D=0{{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 ar,a,ar\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 −BA\dfrac{-B}{A} and the product of the roots is −DA\dfrac{-D}{A} .
So, our equation is in form A3+Bx2+Cx+D=0{{A}^{3}}+B{{x}^{2}}+Cx+D=0. We can write it as
Sum of the roots i.e. ar+a+ar=−BA\dfrac{a}{r}+a+ar=\dfrac{-B}{A} ……………(1)
Product of roots i.e. ar⋅a⋅ar=−DA\dfrac{a}{r}\cdot a\cdot ar=\dfrac{-D}{A} ………………(2)
On putting value in equation (2) where D in out equation is −512-512 and A is 1 we get as
ar⋅a⋅ar=−(−512)1=512\dfrac{a}{r}\cdot a\cdot ar=\dfrac{-\left( -512 \right)}{1}=512
We can also write this equation on cancelling r term as
a3=512{{a}^{3}}=512
Now, we know that the cube of digit 8 is 512. So, we get as
a3=83⇒a=8{{a}^{3}}={{8}^{3}}\Rightarrow a=8 ………………………..(3)
We will now put the value of B i.e. −42-42 and A as 1 in equation (1). We get as
ar+a+ar=−(−42)1=42\dfrac{a}{r}+a+ar=\dfrac{-\left( -42 \right)}{1}=42
Now, on taking LCM of ‘r’ we get as
a+ar+ar2=42ra+ar+a{{r}^{2}}=42r
We will put the value of ‘a’ from equation (3).
8+8r+8r2=42r8+8r+8{{r}^{2}}=42r
On further simplification, we get as
8+8r−42r+8r2=08+8r-42r+8{{r}^{2}}=0
On rearranging the terms, we can write it as
8r2−34r+8=08{{r}^{2}}-34r+8=0
Now, we will take 2 common from the whole equation, we get as
4r2−17r+4=04{{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 (c⋅a)\left( c\cdot a \right) i.e. 16 in this case and on adding we get a middle term i.e. −17-17 . So, we write it as
4r2−16r−r+4=04{{r}^{2}}-16r-r+4=0
Now, we will take 4r common from the first 2 terms and −1-1 from remaining terms. Thus, we get as
4r(r−4)−1(r−4)=04r\left( r-4 \right)-1\left( r-4 \right)=0
(4r−1)(r−4)=0\left( 4r-1 \right)\left( r-4 \right)=0
Thus, on solving we get values of r as 4r−1=0⇒r=144r-1=0\Rightarrow r=\dfrac{1}{4} and r−4=0⇒r=4r-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 ar,a,ar\dfrac{a}{r},a,ar . so, we get as
84,8,8⋅4=2,8,32\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,ar2a,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. a3r3=512{{a}^{3}}{{r}^{3}}=512 . Also, in sum of roots it will be difficult to solve such complex equations i.e. a+ar2+ar3=42a+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.