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Solve x² + 6 = 0

x² + 6 = 0 has no real solutions. The two complex solutions are x = √6 i and x = -√6 i.

x² +x += 0
x2+6=0⇒x=6 i, −6 i (complex)x^2 + 6 = 0 \Rightarrow x = \sqrt{6}\, i,\ -\sqrt{6}\, i\ \text{(complex)}
Discriminant D = -24

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

Graph of y = x² + 6-4-3-2-1123446810121416182022vertexy = x² + 6
Graph of y = x² + 6. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, 6).

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 = 6.

    x2+6=0x^2 + 6 = 0
  2. 2

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

    D=02−4⋅1⋅6=0−24=−24D = 0^2 - 4\cdot1\cdot6 = 0 - 24 = -24
  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±−242x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-24}}{2}
  5. 5

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

    x=6 iorx=−6 ix = \sqrt{6}\, i\quad\text{or}\quad x = -\sqrt{6}\, i
LaTeX source of the formulas on this page
\[ x^2 + 6 = 0\ \Rightarrow\ x = \sqrt{6}\, i,\ -\sqrt{6}\, i \]
\[ x^2 + 6 = 0 \]
\[ D = 0^2 - 4\cdot1\cdot6 = 0 - 24 = -24 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-24}}{2} \]
\[ x = \sqrt{6}\, i\quad\text{or}\quad x = -\sqrt{6}\, i \]

The parabola y = x² + 6

xy
-315
-210
-17
06
17
210
315

Questions

What are the solutions of x² + 6 = 0?

x² + 6 = 0 has no real solutions. The two complex solutions are x = √6 i and x = -√6 i.

Does x² + 6 = 0 factor?

Not over the integers, because the discriminant -24 is negative. Use the quadratic formula.

What is the vertex of y = x² + 6?

(0, 6). 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 = 6.

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