Question

cc.2.2.hs.d.10
use the quadratic formula to solve
the following equations.
note: these will not factor! you have to set = 0 first!
$x^2 + 4x = 2$

$\bigcirc -2 \pm \sqrt{6}$
$\bigcirc -2 \pm 3\sqrt{2}$
$\bigcirc -2 \pm 2 \sqrt{3}$
$\bigcirc 2 \pm i \sqrt{6}$

Explanation:

Step1: Rewrite the equation in standard form

We start with the equation \(x^{2}+4x = 2\). To use the quadratic formula, we need to set the equation to zero. Subtract 2 from both sides:
\(x^{2}+4x - 2=0\)
For a quadratic equation of the form \(ax^{2}+bx + c = 0\), here \(a = 1\), \(b = 4\), and \(c=- 2\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute \(a = 1\), \(b = 4\), and \(c=-2\) into the formula:
First, calculate the discriminant \(\Delta=b^{2}-4ac=(4)^{2}-4\times1\times(-2)=16 + 8=24\)
Then, \(x=\frac{-4\pm\sqrt{24}}{2\times1}=\frac{-4\pm2\sqrt{6}}{2}\)
Simplify the fraction by dividing numerator and denominator by 2:
\(x=-2\pm\sqrt{6}\)

Answer:

\(-2\pm\sqrt{6}\) (corresponding to the option: -2 ± √6)