Question

two solutions for all quadratic equations quick check
solve the quadratic equation $2x^2 + 4x = -12$. which of the following expresses its solutions in the form $a \pm bi$?
(1 point)
\\( -2 \pm 2i\sqrt{5} \\)
\\( 1 \pm i\sqrt{5} \\)
\\( -1 \pm i\sqrt{5} \\)
\\( -1 \pm i\sqrt{7} \\)

Explanation:

Step1: Rewrite the equation in standard form

First, rewrite the quadratic equation \(2x^{2}+4x = - 12\) in the standard form \(ax^{2}+bx + c=0\). Add 12 to both sides: \(2x^{2}+4x + 12=0\). Then divide the entire equation by 2 to simplify: \(x^{2}+2x + 6 = 0\). Here, \(a = 1\), \(b = 2\), \(c=6\).

Step2: Use the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substitute \(a = 1\), \(b = 2\), \(c = 6\) into the formula. First, calculate the discriminant \(\Delta=b^{2}-4ac=(2)^{2}-4\times1\times6=4 - 24=- 20\). Then, \(x=\frac{-2\pm\sqrt{-20}}{2\times1}\). Since \(\sqrt{-20}=\sqrt{20}\times\sqrt{-1}=2\sqrt{5}i\), we have \(x=\frac{-2\pm2\sqrt{5}i}{2}=-1\pm i\sqrt{5}\).

Answer:

\(-1\pm i\sqrt{5}\) (the third option: \(-1\pm i\sqrt{5}\))