Solve 2x² + 4 = 0

2x² + 4 = 0 has no real solutions. The two complex solutions are x = √2 i and x = -√2 i.

x² +x += 0

2x^2 + 4 = 0 \Rightarrow x = \sqrt{2}\, i,\ -\sqrt{2}\, i\ \text{(complex)}

Discriminant D = -32

Change a, b or c (whole numbers, a not 0) and the exact solutions follow.

2x^2 + 4 = 0\ \Rightarrow\ x = \sqrt{2}\, i,\ -\sqrt{2}\, i

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Graph of y = 2x² + 4. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, 4).

How to work it out

1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 2, b = 0, c = 4.
$2x^2 + 4 = 0$
2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
$D = 0^2 - 4\cdot2\cdot4 = 0 - 32 = -32$
3
D is negative, so there are no real solutions; the two solutions are complex numbers.
4
The quadratic formula works for every quadratic:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-32}}{4}$
5
Simplify (pull the square out of the root) to get the solutions:
$x = \sqrt{2}\, i\quad\text{or}\quad x = -\sqrt{2}\, i$

LaTeX source of the formulas on this page

\[ 2x^2 + 4 = 0\ \Rightarrow\ x = \sqrt{2}\, i,\ -\sqrt{2}\, i \]
\[ 2x^2 + 4 = 0 \]
\[ D = 0^2 - 4\cdot2\cdot4 = 0 - 32 = -32 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-32}}{4} \]
\[ x = \sqrt{2}\, i\quad\text{or}\quad x = -\sqrt{2}\, i \]

The parabola y = 2x² + 4

x	y
-3	22
-2	12
-1	6
0	4
1	6
2	12
3	22

Questions

What are the solutions of 2x² + 4 = 0?

2x² + 4 = 0 has no real solutions. The two complex solutions are x = √2 i and x = -√2 i.

Does 2x² + 4 = 0 factor?

Not over the integers, because the discriminant -32 is negative. Use the quadratic formula.

What is the vertex of y = 2x² + 4?

(0, 4). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 0, product = c/a = 2.

Solve 2x² + 4 = 0

How to work it out

The parabola y = 2x² + 4

Questions

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