Question

Solve the quadratic equation 2x2+4x=−12
2
𝑥
2
+
4
𝑥
=
−
12
. Which of the following expresses its solutions in the form a±bi
𝑎
±
𝑏
𝑖

?(1 point)
Responses

1±i5–√
1
±
𝑖
5
1 plus or minus i Start Root 5 End Root

−1±i7–√
−
1
±
𝑖
7
negative 1 plus or minus i Start Root 7 End Root

−2±2i5–√
−
2
±
2
𝑖
5
negative 2 plus or minus 2 i Start Root 5 End Root

−1±i5–√

Answer 1

To solve the quadratic equation \(2x^2 + 4x = -12\), we start by rearranging it into standard form:

\[
2x^2 + 4x + 12 = 0
\]

Next, we'll divide the entire equation by 2 to simplify:

\[
x^2 + 2x + 6 = 0
\]

Now, we can use the quadratic formula to find the solutions. The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, \(a = 1\), \(b = 2\), and \(c = 6\). First, we calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20
\]

Since the discriminant is negative, the solutions will include imaginary numbers. Next, we substitute into the quadratic formula:

\[
x = \frac{-2 \pm \sqrt{-20}}{2 \cdot 1}
\]

We can simplify \(\sqrt{-20}\):

\[
\sqrt{-20} = \sqrt{20} \cdot i = \sqrt{4 \cdot 5} \cdot i = 2\sqrt{5}i
\]

Now, substitute \(\sqrt{-20}\) into the formula:

\[
x = \frac{-2 \pm 2\sqrt{5}i}{2}
\]

This can be further simplified:

\[
x = -1 \pm \sqrt{5}i
\]

Thus, the solutions of the equation \(2x^2 + 4x = -12\) are:

\[
x = -1 \pm i\sqrt{5}
\]

Among the given options, the correct expression for the solutions is:

\[
-1 \pm i\sqrt{5}
\]

So the answer is:

**−1±i5–√**