Question

Solve the equation 6x⁴ – 5x³ – 38x² – 5x + 6 = 0 if it is known that `1/3` is a solution

Answer 1

6x⁴ – 5x³ – 38x² – 5x + 6 = 0

`6(x^2 + 1/x^2) -5(x + 1/x) - 38` = 0

`x + 1/x = y, y^2 - 2 = x^2 + 1/x^2`

6(y² – 2) – 5y – 38 = 0

6y² – 12 – 5y – 38 = 0

6y² – 5y – 50 = 0

6y² – 20y + 15y – 50 = 0

2y(3y – 10) + 5(3y – 10) = 0

(3y – 10)(2y + 5) = 0

3y = 10, 2y = – 5

y = `10/3`, y = `- 5/2`

`x + 1/x = 10/3, x + 1/x = - 5/2`

`(x^2 + 1)/x = 10/3, (x^2 + 1)/x = - 5/2`

3x² + 3 = 10x, 2x² + 2 = – 5x

3x² – 10x + 3 = 0, 2x² + 5x + 2 = 0

(x – 3)(3x – 1) = 0, (2x + 1)(x + 1) = 0

x - 3, x = `1/3` or x = `(- 1)/2`, x = – 2

The roots are  `3, 1/3, -2, (-1)/2`.