Solve 2x² - 4x + 5 = 0
2x² - 4x + 5 = 0 has no real solutions. The two complex solutions are x = (2 + √6 i)/2 and x = (2 - √6 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 = -4, c = 5.
- 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
\[ 2x^2 - 4x + 5 = 0\ \Rightarrow\ x = \frac{2 + \sqrt{6}\, i}{2},\ \frac{2 - \sqrt{6}\, i}{2} \]
\[ 2x^2 - 4x + 5 = 0 \]
\[ D = (-4)^2 - 4\cdot2\cdot5 = 16 - 40 = -24 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{4 \pm \sqrt{-24}}{4} \]
\[ x = \frac{2 + \sqrt{6}\, i}{2}\quad\text{or}\quad x = \frac{2 - \sqrt{6}\, i}{2} \]The parabola y = 2x² - 4x + 5
| x | y |
|---|---|
| -2 | 21 |
| -1 | 11 |
| 0 | 5 |
| 1 | 3 |
| 2 | 5 |
| 3 | 11 |
| 4 | 21 |
Questions
What are the solutions of 2x² - 4x + 5 = 0?
2x² - 4x + 5 = 0 has no real solutions. The two complex solutions are x = (2 + √6 i)/2 and x = (2 - √6 i)/2.
Does 2x² - 4x + 5 = 0 factor?
Not over the integers, because the discriminant -24 is negative. Use the quadratic formula.
What is the vertex of y = 2x² - 4x + 5?
(1, 3). 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 = 5/2.