Question

cc.2.2.hs.d.10
solve for x:
$x^2 - x = -7$
$\bigcirc$ $1/2 \pm(3i/2)\sqrt{3}$
$\bigcirc$ $2 \pm 3i\sqrt{3}$
$\bigcirc$ $1 \pm (3/2)\sqrt{3}$
$\bigcirc$ $1/2 \pm(5i/2)\sqrt{2}$

Explanation:

Step1: Rewrite the equation in standard form

First, we rewrite the given equation \(x^{2}-x = - 7\) into the standard quadratic form \(ax^{2}+bx + c=0\). By adding 7 to both sides, we get \(x^{2}-x + 7=0\). Here, \(a = 1\), \(b=-1\), and \(c = 7\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 1\), \(b=-1\), and \(c = 7\) into the formula.
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times7=1 - 28=- 27\).
Then, \(x=\frac{-(-1)\pm\sqrt{-27}}{2\times1}=\frac{1\pm\sqrt{27}i}{2}\) (since \(\sqrt{-27}=\sqrt{27}\times\sqrt{-1} = 3\sqrt{3}i\)).
Simplify \(\frac{1\pm3\sqrt{3}i}{2}=\frac{1}{2}\pm\frac{3\sqrt{3}i}{2}=\frac{1}{2}\pm\frac{3i}{2}\sqrt{3}\)

Answer:

\(\frac{1}{2}\pm\frac{3i}{2}\sqrt{3}\) (corresponding to the option: \(1/2\pm(3i/2)\sqrt{3}\))