Solve 2x² + 8 = 0

2x² + 8 = 0 has no real solutions. The two complex solutions are x = 2i and x = -2i.

x² +x += 0

2x^2 + 8 = 0 \Rightarrow x = 2i,\ -2i\ \text{(complex)}

Discriminant D = -64

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

2x^2 + 8 = 0\ \Rightarrow\ x = 2i,\ -2i

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Graph of y = 2x² + 8. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, 8).

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 = 8.
$2x^2 + 8 = 0$
2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
$D = 0^2 - 4\cdot2\cdot8 = 0 - 64 = -64$
3
D is negative, so there are no real solutions; the two solutions are complex numbers.
4
The quadratic formula works for every quadratic:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-64}}{4}$
5
Simplify (pull the square out of the root) to get the solutions:
$x = 2i\quad\text{or}\quad x = -2i$

LaTeX source of the formulas on this page

\[ 2x^2 + 8 = 0\ \Rightarrow\ x = 2i,\ -2i \]
\[ 2x^2 + 8 = 0 \]
\[ D = 0^2 - 4\cdot2\cdot8 = 0 - 64 = -64 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-64}}{4} \]
\[ x = 2i\quad\text{or}\quad x = -2i \]

The parabola y = 2x² + 8

x	y
-3	26
-2	16
-1	10
0	8
1	10
2	16
3	26

Questions

What are the solutions of 2x² + 8 = 0?

2x² + 8 = 0 has no real solutions. The two complex solutions are x = 2i and x = -2i.

Does 2x² + 8 = 0 factor?

Not over the integers, because the discriminant -64 is negative. Use the quadratic formula.

What is the vertex of y = 2x² + 8?

(0, 8). 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 = 4.

Solve 2x² + 8 = 0

How to work it out

The parabola y = 2x² + 8

Questions

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