Question

question 4 of 10
what are the solutions of $4x^2 - x + 9 = 0$?
\\(\bigcirc\\) a. $x = \frac{3+i\sqrt{143}}{8}$ or $x = \frac{3-i\sqrt{143}}{8}$
\\(\bigcirc\\) b. $x = \frac{1+6i}{4}$ or $x = \frac{1-6i}{4}$
\\(\bigcirc\\) c. $x = \frac{1+i\sqrt{143}}{8}$ or $x = \frac{1-i\sqrt{143}}{8}$
\\(\bigcirc\\) d. $x = \frac{1+12i}{8}$ or $x = \frac{1-12i}{8}$

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Here, \(a = 4\), \(b=- 1\), \(c = 9\).

Step2: Calculate Discriminant (\(D\))

Discriminant \(D=b^{2}-4ac\).
Substitute \(a = 4\), \(b=-1\), \(c = 9\):
\(D=(-1)^{2}-4\times4\times9=1 - 144=- 143\)

Step3: Apply Quadratic Formula

\(x=\frac{-(-1)\pm\sqrt{-143}}{2\times4}=\frac{1\pm i\sqrt{143}}{8}\) (since \(\sqrt{-143}=i\sqrt{143}\))

Answer:

C. \(x=\frac{1 + i\sqrt{143}}{8}\) or \(x=\frac{1 - i\sqrt{143}}{8}\)