34.1k views
2 votes
Find the solutions of the quadratic equation 3x^2-5x+1=0.

User Gunilla
by
7.6k points

1 Answer

2 votes

Answer:

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)

User Jeff Keller
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories