Solve x² + 2x + 6 = 0
x² + 2x + 6 = 0 has no real solutions. The two complex solutions are x = -1 + √5 i and x = -1 - √5 i.
Change a, b or c (whole numbers, a not 0) and the exact solutions follow.
How to work it out
- 1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 1, b = 2, c = 6.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is negative, so there are no real solutions; the two solutions are complex numbers.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
LaTeX source of the formulas on this page
\[ x^2 + 2x + 6 = 0\ \Rightarrow\ x = -1 + \sqrt{5}\, i,\ -1 - \sqrt{5}\, i \]
\[ x^2 + 2x + 6 = 0 \]
\[ D = 2^2 - 4\cdot1\cdot6 = 4 - 24 = -20 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-20}}{2} \]
\[ x = -1 + \sqrt{5}\, i\quad\text{or}\quad x = -1 - \sqrt{5}\, i \]The parabola y = x² + 2x + 6
| x | y |
|---|---|
| -4 | 14 |
| -3 | 9 |
| -2 | 6 |
| -1 | 5 |
| 0 | 6 |
| 1 | 9 |
| 2 | 14 |
Questions
What are the solutions of x² + 2x + 6 = 0?
x² + 2x + 6 = 0 has no real solutions. The two complex solutions are x = -1 + √5 i and x = -1 - √5 i.
Does x² + 2x + 6 = 0 factor?
Not over the integers, because the discriminant -20 is negative. Use the quadratic formula.
What is the vertex of y = x² + 2x + 6?
(-1, 5). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -2, product = c/a = 6.