Question

solve each equation by using the quadratic formula.
$3x^2 - 13x + 4 = 0$
$circ \frac{1}{3}, 4$
$circ -\frac{1}{3}, -4$
$circ 2, 8$
$circ \frac{13 pm sqrt{217}}{6}$

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For the equation \(3x^{2}-13x + 4 = 0\), we have \(a = 3\), \(b=- 13\), \(c = 4\).

Step2: Calculate the discriminant

The discriminant \(D=b^{2}-4ac\). Substitute \(a = 3\), \(b=-13\), \(c = 4\) into it:
\(D=(-13)^{2}-4\times3\times4=169 - 48=121\)

Step3: Apply Quadratic Formula

\(x=\frac{-(-13)\pm\sqrt{121}}{2\times3}=\frac{13\pm11}{6}\)
We have two cases:

• Case 1: \(x=\frac{13 + 11}{6}=\frac{24}{6}=4\)
• Case 2: \(x=\frac{13-11}{6}=\frac{2}{6}=\frac{1}{3}\)

Answer:

\(\frac{1}{3},4\) (corresponding to the first option: \(\boldsymbol{\frac{1}{3},4}\))