Question

Solve $3{{x}^{2}}-26{{x}^{2}}+52x-24=0$, given that the roots of the equation are in geometric progression.

Answer 1

Hint: Assume the roots of the given cubic equation as $a$, $ar$ and $a{{r}^{2}}$ where ‘a’ is the first term and ‘r’ is the common ratio. Now, compare the given equation with the general form of a cubic equation given as $A{{x}^{3}}+B{{x}^{2}}+Cx+D=0$ and compare the values of A, B, C and D. Use the formula of sum of roots = $\dfrac{-B}{A}$ and product of roots = $\dfrac{-D}{A}$ to find two relations between ‘a’ and ‘r’. Solve the obtained quadratic equation and find the three roots of the equation.

Complete step-by-step answer:
Here we have been provided with the cubic equation $3{{x}^{2}}-26{{x}^{2}}+52x-24=0$ and we are asked to solve this equation with the given condition that the roots are in geometric progression.
Now, since the given equation is a cubic equation so it will have three roots. Let us assume the three roots in G.P as $a$, $ar$ and $a{{r}^{2}}$ where ‘a’ is the first term and ‘r’ is the common ratio. Comparing the equation $3{{x}^{2}}-26{{x}^{2}}+52x-24=0$ with the general form of a cubic equation given as $A{{x}^{3}}+B{{x}^{2}}+Cx+D=0$ we have,
$\Rightarrow $ A = 3, B = -26, C = 52 and D = -24.
We know that the sum of roots of a cubic equation is equal to the ratio $\dfrac{-B}{A}$. So we get,
$\Rightarrow a+ar+a{{r}^{2}}=\dfrac{26}{3}........\left( i \right)$
Also, the product of the roots of a cubic equation is equal to the ratio $\dfrac{-D}{A}$. So we get,
$\begin{align}
  & \Rightarrow a\times ar\times a{{r}^{2}}=\dfrac{24}{3} \\
 & \Rightarrow {{a}^{3}}{{r}^{3}}=8 \\
 & \Rightarrow {{\left( ar \right)}^{3}}=8 \\
\end{align}$
Taking cube root both the sides we get,
$\begin{align}
  & \Rightarrow \left( ar \right)=\sqrt[3]{8} \\
 & \Rightarrow \left( ar \right)=2.............\left( ii \right) \\
\end{align}$
Substituting the value of a from equation (ii) in equation (i) we get,
$\begin{align}
  & \Rightarrow \dfrac{2}{r}+\dfrac{2}{r}\times r+\dfrac{2}{r}\times {{r}^{2}}=\dfrac{26}{3} \\
 & \Rightarrow \dfrac{1}{r}+1+r=\dfrac{13}{3} \\
 & \Rightarrow \dfrac{1+r+{{r}^{2}}}{r}=\dfrac{13}{3} \\
\end{align}$
By cross multiplication we get,
$\begin{align}
  & \Rightarrow 3+3r+3{{r}^{2}}=13r \\
 & \Rightarrow 3{{r}^{2}}-10r+3=0 \\
\end{align}$
Using the middle term split method to factorize the above quadratic equation we get,
$\begin{align}
  & \Rightarrow 3{{r}^{2}}-9r-r+3=0 \\
 & \Rightarrow \left( r-3 \right)\left( 3r-1 \right)=0 \\
\end{align}$
Substituting each term equal to 0 one by one we get,
$\Rightarrow \left( r-3 \right)=0$ or $\left( 3r-1 \right)=0$
$\Rightarrow r=3$ or $r=\dfrac{1}{3}$
Case (i): - Considering $r=3$ we have,
$\Rightarrow a=\dfrac{2}{3}$
Therefore the three roots of the cubic equation will be given as: -
$\Rightarrow a=\dfrac{2}{3}$
$\Rightarrow ar=2$
$\Rightarrow a{{r}^{2}}=6$
Case (ii): - Considering $r=\dfrac{1}{3}$ we have,
$\Rightarrow a=6$
Therefore the three roots of the cubic equation will be given as: -
$\Rightarrow a=6$
$\Rightarrow ar=2$
$\Rightarrow a{{r}^{2}}=\dfrac{2}{3}$
From the above two cases we can conclude that the three roots of the cubic equation in G.P are given as $\dfrac{2}{3}$, $2$ and $6$.

Note: Note that we have obtained two cases for the values of ‘a’ and ‘r’ because in the first case the G.P is increasing as the value of r is greater than 1 while in the second case the G.P is decreasing as the value of r is less than 1. However, there will only be three roots of the cubic equation and that can be written in any order.