Solve 2x² + 2x + 1 = 0
2x² + 2x + 1 = 0 has no real solutions. The two complex solutions are x = (-1 + i)/2 and x = (-1 - i)/2.
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 = 2, b = 2, c = 1.
- 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
\[ 2x^2 + 2x + 1 = 0\ \Rightarrow\ x = \frac{-1 + i}{2},\ \frac{-1 - i}{2} \]
\[ 2x^2 + 2x + 1 = 0 \]
\[ D = 2^2 - 4\cdot2\cdot1 = 4 - 8 = -4 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-4}}{4} \]
\[ x = \frac{-1 + i}{2}\quad\text{or}\quad x = \frac{-1 - i}{2} \]The parabola y = 2x² + 2x + 1
| x | y |
|---|---|
| -3.5 | 18.5 |
| -2.5 | 8.5 |
| -1.5 | 2.5 |
| -0.5 | 0.5 |
| 0.5 | 2.5 |
| 1.5 | 8.5 |
| 2.5 | 18.5 |
Questions
What are the solutions of 2x² + 2x + 1 = 0?
2x² + 2x + 1 = 0 has no real solutions. The two complex solutions are x = (-1 + i)/2 and x = (-1 - i)/2.
Does 2x² + 2x + 1 = 0 factor?
Not over the integers, because the discriminant -4 is negative. Use the quadratic formula.
What is the vertex of y = 2x² + 2x + 1?
(-1/2, 1/2). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -1, product = c/a = 1/2.