Solve -x² + 2x - 5 = 0
-x² + 2x - 5 = 0 has no real solutions. The two complex solutions are x = 1 + 2i and x = 1 - 2i.
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 = -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
\[ -x^2 + 2x - 5 = 0\ \Rightarrow\ x = 1 + 2i,\ 1 - 2i \]
\[ -x^2 + 2x - 5 = 0 \]
\[ D = 2^2 - 4\cdot(-1)\cdot(-5) = 4 - 20 = -16 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-16}}{-2} \]
\[ x = 1 + 2i\quad\text{or}\quad x = 1 - 2i \]The parabola y = -x² + 2x - 5
| x | y |
|---|---|
| -2 | -13 |
| -1 | -8 |
| 0 | -5 |
| 1 | -4 |
| 2 | -5 |
| 3 | -8 |
| 4 | -13 |
Questions
What are the solutions of -x² + 2x - 5 = 0?
-x² + 2x - 5 = 0 has no real solutions. The two complex solutions are x = 1 + 2i and x = 1 - 2i.
Does -x² + 2x - 5 = 0 factor?
Not over the integers, because the discriminant -16 is negative. Use the quadratic formula.
What is the vertex of y = -x² + 2x - 5?
(1, -4). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 2, product = c/a = 5.