Question

Solve the equation $ 27{{x}^{4}}-195{{x}^{3}}+494{{x}^{2}}-520x+192=0 $, the roots being in geometric progression.

Answer 1

Hint: Take $a,b,c,d$ as the four roots of the given fourth degree equation. Find $ad,bc$ using product of roots formula. Formulate two quadratic equations one in $a,d$ other in $b,c$. Multiply respective sides of both the equations to get an equation in fourth degree and finally equate left hand side of obtained and the given equation to find $a,b,c,d$.\[\]

Complete step-by-step answer:
We know that the product of roots of a polynomial of degree $n$ , ${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...+{{a}_{0}}=0$ is given $\Pi ={{\left( -1 \right)}^{n}}\dfrac{{{a}_{0}}}{{{a}_{n}}}$. \[\]
The given equation is ,
\[\begin{align}
  & p\left( x \right)=27{{x}^{4}}-195{{x}^{3}}+494{{x}^{2}}-520x+192=0. \\
 & \Rightarrow {{x}^{4}}-\dfrac{195}{27}{{x}^{3}}+\dfrac{494}{27}{{x}^{2}}-\dfrac{520}{27}x+\dfrac{192}{27}=0 \\
 & \Rightarrow {{x}^{4}}-\dfrac{65}{9}{{x}^{3}}+\dfrac{494}{27}{{x}^{2}}-\dfrac{520}{27}x+\dfrac{64}{9}=0...(1) \\
\end{align}\]
We denoted the expression at the left hand side as $p\left( x \right)$ . The given equation is of fourth degree. So it will have roots and as they are in geometric progression all of them are real. Let the roots of the equation be $a,b,c,d.$ As it is given in the question the roots are in geometric progression(GP) then
\[\dfrac{b}{a}=\dfrac{d}{c}\Rightarrow ad=bc\].
We use the product of roots to find

\[\begin{align}
  & abcd={{\left( -1 \right)}^{4}}\dfrac{64}{9} \\
 & \Rightarrow \left( ad \right)\left( bc \right)=\dfrac{64}{9} \\
 & \Rightarrow {{\left( ad \right)}^{2}}=\dfrac{64}{9} \\
 & \Rightarrow ad=bc=\dfrac{8}{3} \\
\end{align}\]

We only take the positive roots now. We formulate a quadratic equation with roots $a$ and $d$ and get,
\[\begin{align}
  & \left( x-a \right)\left( x-d \right)=0 \\
 & \Rightarrow {{x}^{2}}-\left( a+d \right)x+ad=0 \\
 & \Rightarrow {{x}^{2}}-Mx+\dfrac{8}{3}=0....(2) \\
\end{align}\]
Where $M=a+d$, We formulate another quadratic equation with roots $b$ and $c$ and get,
\[\begin{align}
  & \left( x-b \right)\left( x-c \right)=0 \\
 & \Rightarrow {{x}^{2}}-\left( b+c \right)x+bc=0 \\
 & \Rightarrow {{x}^{2}}-Nx+\dfrac{8}{3}=0....(3) \\
\end{align}\]
Where $N=b+c$. We multiply respective sides equations of equation (2) and (3). We get,

\[{{x}^{4}}-\left( M+N \right){{x}^{3}}+\left( MN+\dfrac{16}{3} \right){{x}^{2}}-\dfrac{8}{3}\left( M+N \right)x+\dfrac{64}{9}=0...(4)\]

We equate the coefficients of terms in $x$ and ${{x}^{2}}$ appearing in the left hand sides of equation (1) and (4). We have,
\[\begin{align}
  & M+N=\dfrac{520}{27}\times \left( \dfrac{-3}{8} \right)=-\dfrac{65}{9} \\
 & MN=-\dfrac{494}{27}-\dfrac{16}{3}=\dfrac{350}{27} \\
\end{align}\]
We eliminate $N$ from above equation by putting $N=\dfrac{350}{27M}$ in $M+N=-\dfrac{65}{9}$ and get,
\[ 27{{M}^{2}}+195M+350=0 \]
We solve the above quadratic equation by splitting the middle term method. We factorize $27\times 350=9450$ and find two factors whose sum is 195 which are 105 and 90.
\[ \begin{align}
  & 27{{M}^{2}}+105M+90M+350=0 \\
 & \Rightarrow \left( 3M+10 \right)\left( 9M+35 \right)=0 \\
\end{align}\]
So we obtain $M=\dfrac{-35}{9}$ and $N=\dfrac{350}{27M}=\dfrac{-10}{3}$(We also can take $M=\dfrac{-10}{3}$). We put $M$ in equation (2) and $N$ in equation (3) and get ,
\[\begin{align}
  & {{x}^{2}}+\dfrac{35}{9}x+\dfrac{8}{3}=0\Rightarrow 9{{x}^{2}}+35x+27=0\left( \text{from}\left( 2 \right) \right) \\
 & {{x}^{2}}+\dfrac{10}{3}x+\dfrac{8}{3}=0\Rightarrow 3{{x}^{2}}+10x+8=0\left( \text{from}\left( 3 \right) \right) \\
\end{align}\]
We solve the above pair of quadratic equation by splitting the middle term method.
\[\begin{align}
  & 9{{x}^{2}}+35x+27=0\Rightarrow \left( x+3 \right)\left( 9x+8 \right)=0 \\
 & 3{{x}^{2}}+10x+8=0\Rightarrow \left( 3x+4 \right)\left( x+2 \right)=0 \\
\end{align}\]
So the roots of the equation are $-3,\dfrac{-8}{9},-2,\dfrac{-4}{3}.$ We verify that their product is $\dfrac{-64}{9}$ and arrange them in geometric progression $-3,-2,\dfrac{-4}{3},\dfrac{-8}{9}.$\[\]

Note: We need to take care of the confusion of relation between four terms of GP from AP which is $a+d=b+c.$ We can also solve the problem with four generalized terms of GP that is $a,ar,a{{r}^{2}},a{{r}^{3}}$. We should also note that we can solve the problem taking $ad=\dfrac{-8}{3}$ . The sum roots of a polynomial of degree $n$ is $-\dfrac{{{a}_{n-1}}}{{{a}_{0}}}$.