Question

question 2 of 10
use the quadratic formula to find both solutions to the quadratic equation
given below.

$3x^2 - 5x + 1 = 0$

a. $x = \frac{5 + \sqrt{37}}{6}$

b. $x = \frac{-5 + \sqrt{13}}{6}$

c. $x = \frac{5 - \sqrt{13}}{6}$

d. $x = \frac{5 + \sqrt{13}}{6}$

e. $x = \frac{5 - \sqrt{37}}{6}$

f. $x = \frac{-5 - \sqrt{13}}{6}$

Explanation:

Step1: Recall quadratic formula

The quadratic formula for a quadratic equation \(ax^2 + bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). For the equation \(3x^2-5x + 1=0\), we have \(a = 3\), \(b=- 5\), \(c = 1\).

Step2: Calculate discriminant

First, calculate the discriminant \(D=b^2-4ac\). Substitute \(a = 3\), \(b=-5\), \(c = 1\) into the formula:
\(D=(-5)^2-4\times3\times1=25 - 12=13\)

Step3: Apply quadratic formula

Substitute \(a = 3\), \(b=-5\), \(D = 13\) into the quadratic formula:
\(x=\frac{-(-5)\pm\sqrt{13}}{2\times3}=\frac{5\pm\sqrt{13}}{6}\)
So the two solutions are \(x=\frac{5 + \sqrt{13}}{6}\) and \(x=\frac{5-\sqrt{13}}{6}\)

Answer:

C. \(x=\frac{5-\sqrt{13}}{6}\), D. \(x=\frac{5+\sqrt{13}}{6}\)