Question

Find the value of k in the equation ${{x}^{3}}-6{{x}^{2}}+kx+64=0$ , if it is known that the roots of the equation are in geometric progression.
(A) 24
(B) 16
(C) -16
(D) -24

Answer 1

Assume the assume the first term of the geometric progression be a and the common ratio be r. The roots of the cubic equation ${{x}^{3}}-6{{x}^{2}}+kx+64=0$ are in geometric progression. The roots of the cubic equation ${{x}^{3}}-6{{x}^{2}}+kx+64=0$ are $a$ , $ar$ , and $a{{r}^{2}}$ . Take $a$ , $ar$ , and $a{{r}^{2}}$ as roots and then use these three formulas, the sum of all roots = $\dfrac{-b}{a}$ , the sum of the products of the roots taken two at a time = $\dfrac{c}{a}$, and the product of all roots = $\dfrac{-d}{a}$ . Now, solve these three equations and get the value of k.

Complete step by step answer:
According to the question, we have the cubic equation, ${{x}^{3}}-6{{x}^{2}}+kx+64=0$ and the roots of this equation are in geometric progression. Since, the equation is cubic so, we have three roots.
First of all, let us assume the first term of the geometric progression be a and the common ratio be r.
The terms of the geometric progression are the roots of the cubic equation.
Now, the first root of the cubic equation = $a$ …………………….…(1)
The second root of the cubic equation = $ar$ ………………..(2)
The third root of the cubic equation = $a{{r}^{2}}$ ………………..(3)
We know the formulas for the cubic equation of the form
$a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ ………………………..(4)
The sum of all roots = $\dfrac{-b}{a}$ ………………….(5)
The sum of the products of the roots taken two at a time = $\dfrac{c}{a}$ ………………………….(6)
The product of all roots = $\dfrac{-d}{a}$ …………………………(7)
We have the cubic equation ${{x}^{3}}-6{{x}^{2}}+kx+64=0$ ……………………………(8)
Now, comparing equation (4) and equation (7), we get
$a\text{ }=\text{ }1$ ………………………….(9)
$b=-6$ …………………………..(10)
$c=k$ …………………………….(11)
$d=64$ …………………………..(12)
From equation (7), equation (9), and equation (12), we get
The product of all roots = $\dfrac{-64}{1}=-64$ …………………….(13)
Now, from equation (1), equation (2), equation (3) and equation (13), we get

& a\left( ar \right)\left( a{{r}^{2}} \right)=-64 \\\ & \Rightarrow {{\left( ar \right)}^{3}}=-64 \\\ & \Rightarrow ar=-4 \\\ \end{aligned}$$ $$\Rightarrow r=\dfrac{-4}{a}$$ ……………………………(14) From equation (5), equation (9), and equation (10), we get The sum of all roots = $$\dfrac{-\left( -6 \right)}{1}=6$$ …………………….(15) Now, from equation (1), equation (2), equation (3) and equation (15), we get $$a+ar+a{{r}^{2}}=-64$$ …………………………(16) Now from equation (14) and equation (16), we get $$\begin{aligned} & a+a\left( \dfrac{-4}{a} \right)+a{{\left( \dfrac{-4}{a} \right)}^{2}}=6 \\\ & \Rightarrow a-4+a\left( \dfrac{16}{{{a}^{2}}} \right)=6 \\\ & \Rightarrow a+\dfrac{16}{a}=6+4 \\\ & \Rightarrow a+\dfrac{16}{a}=10 \\\ & \Rightarrow \dfrac{{{a}^{2}}+16}{a}=10 \\\ & \Rightarrow {{a}^{2}}+16=10a \\\ & \Rightarrow {{a}^{2}}-10a+16=0 \\\ & \Rightarrow {{a}^{2}}-8a-2a+16=0 \\\ & \Rightarrow a\left( a-8 \right)-2\left( a-8 \right)=0 \\\ & \Rightarrow \left( a-2 \right)\left( a-8 \right)=0 \\\ \end{aligned}$$ $$a=2$$ or $$a=8$$ Let us take $$a=2$$ ………………..(17) Putting $$a=2$$ in the equation (14), we get $$\Rightarrow r=\dfrac{-4}{2}=-2$$ ………………………….(18) Now, from equation (6), equation (9), and equation (11), we get The sum of the products of the roots taken two at a time = $$\dfrac{k}{1}$$ ……………………………(19) From equation (1), equation (2), equation (3) and equation (19), we get $$a.ar+ar.a{{r}^{2}}+a{{r}^{2}}.a=k$$ …………………..(20) Now, putting the values of a from equation (17) and r from equation (18), in equation (20), we get $$\begin{aligned} & 2\times 2\left( -2 \right)+2\left( -2 \right).2{{\left( -2 \right)}^{2}}+2{{\left( -2 \right)}^{2}}.2=k \\\ & \Rightarrow 4\left( -2 \right)+16\left( -2 \right)+16=k \\\ & \Rightarrow -8-32+16=k \\\ & \Rightarrow -24=k \\\ \end{aligned}$$ Therefore, the value of k is -24 …………………………………………….(21) Let us take $$a=8$$ ………………..(22) Putting $$a=8$$ in the equation (14), we get $$\Rightarrow r=\dfrac{-4}{8}=-\dfrac{1}{2}$$ ………………………….(23) Now, from equation (6), equation (9), and equation (11), we get The sum of the products of the roots taken two at a time = $$\dfrac{k}{1}$$ ……………………………(24) From equation (1), equation (2), equation (3) and equation (19), we get $$a.ar+ar.a{{r}^{2}}+a{{r}^{2}}.a=k$$ …………………..(25) Now, putting the values of a from equation (17) and r from equation (18), in equation (20), we get $$\begin{aligned} & 2\times 2\left( -\dfrac{1}{2} \right)+2\left( -\dfrac{1}{2} \right).2{{\left( -\dfrac{1}{2} \right)}^{2}}+2{{\left( -\dfrac{1}{2} \right)}^{2}}.2=k \\\ & \Rightarrow \left( -2 \right)+\left( -1 \right)2\times \left( \dfrac{1}{4} \right)+1=k \\\ & \Rightarrow -2-\dfrac{1}{2}+1=k \\\ & \Rightarrow -\dfrac{3}{2}=k \\\ \end{aligned}$$ Therefore, the value of k is $$\dfrac{-3}{2}$$ …………………………………………….(26) From equation (21) and equation (26), we have two values of k. The value of k that is, $$-\dfrac{3}{2}=k$$ is not given in the option. Only $$k=-24$$ is given in the option. **So, the correct answer is “Option D”.** **Note:** Since the roots of the cubic equation $${{x}^{3}}-6{{x}^{2}}+kx+64=0$$ are in geometric progression so, one might take the summation of roots as $$\dfrac{a}{1-r}$$ . This is wrong because the summation of the infinite geometric progression is $$\dfrac{a}{1-r}$$ and here we don’t have infinite geometric progression. So, we cannot use the formula of summation of infinite geometric series here.