Solve 2x² + 5 = 0
2x² + 5 = 0 has no real solutions. The two complex solutions are x = (√10 i)/2 and x = (-√10 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 = 0, 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
\[ 2x^2 + 5 = 0\ \Rightarrow\ x = \frac{\sqrt{10}\, i}{2},\ \frac{-\sqrt{10}\, i}{2} \]
\[ 2x^2 + 5 = 0 \]
\[ D = 0^2 - 4\cdot2\cdot5 = 0 - 40 = -40 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-40}}{4} \]
\[ x = \frac{\sqrt{10}\, i}{2}\quad\text{or}\quad x = \frac{-\sqrt{10}\, i}{2} \]The parabola y = 2x² + 5
| x | y |
|---|---|
| -3 | 23 |
| -2 | 13 |
| -1 | 7 |
| 0 | 5 |
| 1 | 7 |
| 2 | 13 |
| 3 | 23 |
Questions
What are the solutions of 2x² + 5 = 0?
2x² + 5 = 0 has no real solutions. The two complex solutions are x = (√10 i)/2 and x = (-√10 i)/2.
Does 2x² + 5 = 0 factor?
Not over the integers, because the discriminant -40 is negative. Use the quadratic formula.
What is the vertex of y = 2x² + 5?
(0, 5). 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 = 5/2.