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Solve -x² - 2 = 0

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

x² +x += 0
−x2−2=0⇒x=2 i, −2 i (complex)-x^2 - 2 = 0 \Rightarrow x = \sqrt{2}\, i,\ -\sqrt{2}\, i\ \text{(complex)}
Discriminant D = -8

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

Graph of y = -x² - 2-4-3-2-11234-18-16-14-12-10-8-6-4-2vertexy = -x² - 2
Graph of y = -x² - 2. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, -2).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -1, b = 0, c = -2.

    −x2−2=0-x^2 - 2 = 0
  2. 2

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

    D=02−4⋅(−1)⋅(−2)=0−8=−8D = 0^2 - 4\cdot(-1)\cdot(-2) = 0 - 8 = -8
  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±−8−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-8}}{-2}
  5. 5

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

    x=2 iorx=−2 ix = \sqrt{2}\, i\quad\text{or}\quad x = -\sqrt{2}\, i
LaTeX source of the formulas on this page
\[ -x^2 - 2 = 0\ \Rightarrow\ x = \sqrt{2}\, i,\ -\sqrt{2}\, i \]
\[ -x^2 - 2 = 0 \]
\[ D = 0^2 - 4\cdot(-1)\cdot(-2) = 0 - 8 = -8 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-8}}{-2} \]
\[ x = \sqrt{2}\, i\quad\text{or}\quad x = -\sqrt{2}\, i \]

The parabola y = -x² - 2

xy
-3-11
-2-6
-1-3
0-2
1-3
2-6
3-11

Questions

What are the solutions of -x² - 2 = 0?

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

Does -x² - 2 = 0 factor?

Not over the integers, because the discriminant -8 is negative. Use the quadratic formula.

What is the vertex of y = -x² - 2?

(0, -2). 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 = 2.

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