Question

1. what are the complex solutions to the following equation?

$2x^{2}+4x + 6 = 0$
$\bigcirc -1\pm\sqrt{2}$
$\bigcirc -1\pm i\sqrt{2}$
$\bigcirc 1\pm i\sqrt{2}$
$\bigcirc -3\pm i\sqrt{2}$

Explanation:

Step1: Identify the 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, $a = 2$, $b = 4$, $c = 6$.

Step2: Calculate the discriminant

The discriminant $D = b^2 - 4ac$. Substituting the values, we get $D=(4)^2 - 4\times2\times6 = 16 - 48 = -32$.

Step3: Simplify the square root of the discriminant

Since $D=-32$, $\sqrt{D}=\sqrt{-32}=\sqrt{32}\times\sqrt{-1}=4\sqrt{2}i$ (because $\sqrt{-1}=i$).

Step4: Apply the quadratic formula

Now, $x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-4\pm4\sqrt{2}i}{2\times2}=\frac{-4\pm4\sqrt{2}i}{4}$.

Step5: Simplify the expression

Divide numerator and denominator by 4: $x=-1\pm i\sqrt{2}$.

Answer:

B. $-1\pm i\sqrt{2}$