Question

Find the solutions of the quadratic equation 3x^2-5x+1=0.

Answer 1

The solutions of the quadratic equation are
`x_(1) = (5 + √(13))/(6), x_(2) = (5 - √(13))/(6)`

Explanation:

This is a second order polynomial, and we can find it's roots by the Bhaskara formula.

Explanation of the bhaskara formula:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = (x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

For this problem, we have to find
`x_(1) \text{and} x_(2)`.

The polynomial is
`3x^(2) - 5x +1`, so a = 3, b = -5, c = 1.

Solution

`\bigtriangleup = b^(2) - 4ac = (-5)^(2) - 4*3*1 = 25 - 12 = 13`

`x_(1) = (-b + √(\bigtriangleup))/(2*a) = (-(-5) + √(13))/(2*3) = (5 + √(13))/(6)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a) = (-(-5) - √(13))/(2*3) = (5 - √(13))/(6)`

The solutions of the quadratic equation are
`x_(1) = (5 + √(13))/(6), x_(2) = (5 - √(13))/(6)`