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Solve -2x² - 8 = 0

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

x² +x += 0
−2x2−8=0⇒x=2i, −2i (complex)-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.

Graph of y = -2x² - 8-4-3-2-11234-40-35-30-25-20-15-10-5vertexy = -2x² - 8
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. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -2, b = 0, c = -8.

    −2x2−8=0-2x^2 - 8 = 0
  2. 2

    Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.

    D=02−4⋅(−2)⋅(−8)=0−64=−64D = 0^2 - 4\cdot(-2)\cdot(-8) = 0 - 64 = -64
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=0±−64−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-64}}{-4}
  5. 5

    Simplify (pull the square out of the root) to get the solutions:

    x=2iorx=−2ix = 2i\quad\text{or}\quad x = -2i
LaTeX source of the formulas on this page
\[ -2x^2 - 8 = 0\ \Rightarrow\ x = 2i,\ -2i \]
\[ -2x^2 - 8 = 0 \]
\[ D = 0^2 - 4\cdot(-2)\cdot(-8) = 0 - 64 = -64 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-64}}{-4} \]
\[ x = 2i\quad\text{or}\quad x = -2i \]

The parabola y = -2x² - 8

xy
-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 downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 0, product = c/a = 4.

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