Question

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

Answer 1

Answer:

The solutions of the quadratic equation are $x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}$

Step-by-step 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:

$ax^{2} + bx + c, a\neq0$.

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:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

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

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

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) + \sqrt{13}}{2*3} = \frac{5 + \sqrt{13}}{6}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a} = \frac{-(-5) - \sqrt{13}}{2*3} = \frac{5 - \sqrt{13}}{6}$

The solutions of the quadratic equation are $x_{1} = \frac{5 + \sqrt{13}}{6}, x_{2} = \frac{5 - \sqrt{13}}{6}$