Question

solve the equation 3x^4-40x^3+130x^2-120x+27=0 whose roots are in GP. [Hint: Assume that the roots to be a÷r^3,a÷r,ar,ar^3]​

Answer 1

Given:

The equation 3x^4-40x^3+130x^2-120x+27=0

To find:

Solve the equation 3x^4-40x^3+130x^2-120x+27=0 whose roots are in GP.

Solution:

From the given information, we have the data as follows.

The equation 3x⁴ - 40x³ + 130x² - 120x + 27 = 0

We are asked to find the value of the roots of the given equation.

So, we get,

(x - 1) (x - 3) (3x - 1) (x - 9) = 0

Continue the further calculation.

(x - 1) = 0 ⇒ x = 1

(x - 3) = 0 ⇒ x = 3

(3x - 1) = 0 ⇒ x = 1/3

(x - 9) = 0 ⇒ x = 9

Therefore, the roots of the equation 3x⁴ - 40x³ + 130x² - 120x + 27 = 0 are 1, 3, 1/3 and 9.