Question

question 9 of 10
use the quadratic formula to find both solutions to the quadratic equation
given below.
$3x^2 - x + 4 = 0$

a. $x = \frac{1 - \sqrt{12}}{2}$
b. $x = \frac{1 - \sqrt{48}}{6}$
c. $x = \frac{1 - \sqrt{-47}}{6}$
d. $x = \frac{1 + \sqrt{-47}}{6}$
e. $x = \frac{1 + \sqrt{48}}{6}$
f. $x = \frac{1 + \sqrt{12}}{2}$

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 \(3x^{2}-x + 4=0\), we have \(a = 3\), \(b=- 1\), and \(c = 4\).

Step2: Calculate discriminant \(D=b^{2}-4ac\)

Substitute \(a = 3\), \(b=-1\), \(c = 4\) into the discriminant formula:
\(D=(-1)^{2}-4\times3\times4=1 - 48=-47\)

Step3: Substitute into quadratic formula

Substitute \(a = 3\), \(b=-1\), and \(D=-47\) into the quadratic formula:
\(x=\frac{-(-1)\pm\sqrt{-47}}{2\times3}=\frac{1\pm\sqrt{-47}}{6}\)
So the two solutions are \(x=\frac{1+\sqrt{-47}}{6}\) and \(x=\frac{1 - \sqrt{-47}}{6}\)

Answer:

C. \(x=\frac{1-\sqrt{-47}}{6}\), D. \(x=\frac{1+\sqrt{-47}}{6}\)