Question

Solve the equation ${x^3} - 7{x^2} + 14x - 8 = 0$, given that the roots are in geometric progression.

Answer 1

To solve this question first we have to assume the one root and then go assume another root according to the geometric progression. And apply all the conditions of sum of roots and product of roots and from those two equations, we will find the value of terms of geometric progression and from that find the roots of the given equation.

Complete step-by-step solution:
Given,
A cubic equation
${x^3} - 7{x^2} + 14x - 8 = 0$
Given condition: all the roots are in geometric progression.
To find,
The roots of the cubic equation ${x^3} - 7{x^2} + 14x - 8 = 0$.
In order to solve the question first, we have to assume the first root of the equation.
Let, the first root of the equation is $\dfrac{a}{r}$.
Here, the $a$ is the first term of the geometric progression.
And $r$ is the second term of the geometric progression.
According to the condition. All the roots are
First root: $\dfrac{a}{r}$
Second root: $a$
Third root: $ar$
Applying all the conditions of the sum of roots and product of the roots.
$\text{Sum of the roots} = - \dfrac{{\text{coefficient of } {x^2}}}{{\text{coefficient of } {x^3} }}$
$\dfrac{a}{r} + a + ar = - \dfrac{b}{a}$ ………………(i)
$\text{Product of the roots} = - \dfrac{\text{constant term}}{{\text{coefficient of} {x^3} }}$
$\dfrac{a}{r} \times a \times ar = - \dfrac{d}{a}$ ……………(ii)
On putting all the values of $d$ and $a$ in equation (ii)
$\dfrac{a}{r} \times a \times ar = - \dfrac{{ - 8}}{1}$
$r$ in nominator is canceled by $r$ in denominator
${a^3} = 8$
On taking cube root both side we get,
$a = 2$
On putting all the values in the equation one.
$\dfrac{a}{r} + a + ar = - \dfrac{b}{a}$
$\Rightarrow \dfrac{2}{r} + 2 + 2r = - \dfrac{{ - 7}}{1}$
On further solving
$\dfrac{2}{r} + 2r = 7 - 2$
Taking LCM on denominator.
$\dfrac{{2{r^2} + 2}}{r} = 5$
Taking $r$ to the other side of the equal
$2{r^2} + 2 = 5r$
On arranging this equation.
$\Rightarrow 2{r^2} - 5r + 2 = 0$
This is the quadratic equation now we have to solve this equation in order to find the value of r.
$2{r^2} - 4r - r + 2 = 0$
On taking common after factorization.
$2r(r - 2) - 1(r - 2) = 0$
$\Rightarrow (2r - 1)(r - 2) = 0$
From here we get the two values of $r$.
$\Rightarrow r = \dfrac{1}{2}$ and
$\Rightarrow r = 2$
If we put both the values then we get same roots of the equation
Checking by $r = \dfrac{1}{2}$
Roots are $\dfrac{a}{r}, a, ar$
On putting the values of a and r the roots are $4,2,1$
And by checking by $r = 2$
Roots are $\dfrac{a}{r},a,ar$
On putting the values of a and r the roots are $1,2,4$
So the roots of the equation ${x^3} - 7{x^2} + 14x - 8 = 0$ are $1,2,4$.

Note: To find this type of question we have to assume the first root and make another root according to the condition. Then all roots are assumed in only two variables and apply all the conditions of sum and product of that equation and find all the known variables that we assume in the starting. From that value find the value of roots of the equation.