Question

Solve the equation: $32{x^3} - 48{x^2} + 22x - 3 = 0$ , the roots being in arithmetic progression.

Answer 1

Hint: Since the roots are in arithmetical progression and the highest degree of the equation is 3 then the roots can be assumed to be a-d, a and a+d where a is the starting number and d is the common difference of this arithmetic progression. Then we will find the sum of the roots and product of the roots as per the cubic equation. From there we will get two equations. By solving the equations, we will get the value of a and d. Now just using these values, we will get the final answer.

Complete step-by-step answer:
$32{x^3} - 48{x^2} + 22x - 3 = 0$
As this is a cubic-equation then it has exactly 3 roots.
According to the question the roots are in arithmetic progression,
$\therefore $ Let the roots be a-d, a and a+d.
Where a is the starting number and d is the common difference.
The general cubic-equation is of the form $A{x^3} + B{x^2} + Cx + D$
On comparing it with the equation $32{x^3} - 48{x^2} + 22x - 3 = 0$ we get,
$A = 32$
$B = - 48$
$C = 22$
$D = - 3$
As it is known that,
\[Sum{\text{ }}of{\text{ }}roots{\text{ }}of{\text{ }}cubic{\text{ - }}equation = - \dfrac{B}{A}\]
$\because {\text{ the roots are a - d, a, a + d}}$
$\therefore $ On putting the values,
$a - d + a + a + d = - \left( {\dfrac{{ - 48}}{{32}}} \right)$
On solving further,
$ \Rightarrow 3a = - \left( {\dfrac{{ - 48}}{{32}}} \right)$
On simplifying,
$a = \dfrac{1}{2}$
Also,
\[Product{\text{ }}of{\text{ }}roots{\text{ }}of{\text{ cubic - }}equation = - \dfrac{D}{A}\]
On putting the values,
$\left( {a - d} \right)\left( a \right)\left( {a + d} \right) = - \left( {\dfrac{{ - 3}}{{32}}} \right)$
$ \Rightarrow a\left( {{a^2} - {d^2}} \right) = \left( {\dfrac{3}{{32}}} \right)$
On putting the value of a,
$ \Rightarrow \dfrac{1}{2}\left( {{{\left( {\dfrac{1}{2}} \right)}^2} - {d^2}} \right) = \left( {\dfrac{3}{{32}}} \right)$
On dividing $\dfrac{3}{2}$ both sides,
$ \Rightarrow \left( {\dfrac{1}{4} - {d^2}} \right) = \left( {\dfrac{3}{{32}}} \right) \times \dfrac{2}{1}$
On simplifying further,
$\left( {\dfrac{1}{4} - {d^2}} \right) = \left( {\dfrac{3}{{16}}} \right)$
On subtracting $\dfrac{1}{4}$ both sides,
$ - {d^2} = \dfrac{3}{{16}} - \dfrac{1}{4}$
On taking LCM and simplifying,
$ - {d^2} = - \dfrac{1}{{16}}$
Multiplying -1 both sides,
${d^2} = \dfrac{1}{{16}}$
Takin square root both sides,
$ \Rightarrow \sqrt {{d^2}} = \sqrt {\dfrac{1}{{16}}} $
$ \Rightarrow d = \pm \dfrac{1}{4}$
$The{\text{ roots are }}\dfrac{1}{2} - \dfrac{1}{4}{\text{, }}\dfrac{1}{2}{\text{ and }}\dfrac{1}{2} + \dfrac{1}{4}$
$\therefore {\text{ }}The{\text{ roots are }}\dfrac{1}{4}{\text{, }}\dfrac{1}{2}{\text{ and }}\dfrac{3}{4}$
$\therefore {\text{ The final answer is t}}he{\text{ roots of the given cubic equation are }}\dfrac{1}{4}{\text{, }}\dfrac{1}{2}{\text{ and }}\dfrac{3}{4}$

Note: To solve this problem one must know the general form of the cubic-equation. The equation by which the roots and the coefficients of the equation are related is also to be known. Calculations should be done carefully to avoid any mistake. Always try to solve the question step by step. One can also assume the roots to be a-2d, a and a+2d, the answer would not change.