Question

so no matter what method you chose (completing
t the same solutions!
simplest radical form.
b) $3x^2 + 4x - 1 = 0$
$a=3$ $b=4$ $c=-1$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-(4) \pm \sqrt{(4)^2 - 4(3)(-1)}}{2(3)}$
$= \frac{-4 \pm \sqrt{28}}{6} = \frac{-4 \pm \sqrt{4 \cdot 7}}{6}$
$= \frac{-4 \pm 2\sqrt{7}}{6}$
$\frac{-4}{6} \pm \frac{2\sqrt{7}}{6} = -\frac{2}{3} \pm \frac{\sqrt{7}}{3}$

Explanation:

Step1: Identify coefficients

For quadratic equation \(3x^2 + 4x - 1 = 0\), \(a = 3\), \(b = 4\), \(c = -1\).

Step2: Apply quadratic formula

Quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Substitute \(a\), \(b\), \(c\):
\(x=\frac{-4\pm\sqrt{4^2 - 4(3)(-1)}}{2(3)}\)

Step3: Simplify discriminant

Calculate \(b^2 - 4ac = 16 + 12 = 28\), so \(x=\frac{-4\pm\sqrt{28}}{6}\).

Step4: Simplify radical

\(\sqrt{28}=\sqrt{4\times7}=2\sqrt{7}\), so \(x=\frac{-4\pm2\sqrt{7}}{6}\).

Step5: Reduce fraction

Divide numerator and denominator by 2: \(x=\frac{-2\pm\sqrt{7}}{3}\).

Answer:

\(x = \frac{-2 + \sqrt{7}}{3}\) or \(x = \frac{-2 - \sqrt{7}}{3}\)