Solve x² + x + 4 = 0
x² + x + 4 = 0 has no real solutions. The two complex solutions are x = (-1 + √15 i)/2 and x = (-1 - √15 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 = 1, b = 1, c = 4.
- 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
\[ x^2 + x + 4 = 0\ \Rightarrow\ x = \frac{-1 + \sqrt{15}\, i}{2},\ \frac{-1 - \sqrt{15}\, i}{2} \]
\[ x^2 + x + 4 = 0 \]
\[ D = 1^2 - 4\cdot1\cdot4 = 1 - 16 = -15 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{-15}}{2} \]
\[ x = \frac{-1 + \sqrt{15}\, i}{2}\quad\text{or}\quad x = \frac{-1 - \sqrt{15}\, i}{2} \]The parabola y = x² + x + 4
| x | y |
|---|---|
| -3.5 | 12.75 |
| -2.5 | 7.75 |
| -1.5 | 4.75 |
| -0.5 | 3.75 |
| 0.5 | 4.75 |
| 1.5 | 7.75 |
| 2.5 | 12.75 |
Questions
What are the solutions of x² + x + 4 = 0?
x² + x + 4 = 0 has no real solutions. The two complex solutions are x = (-1 + √15 i)/2 and x = (-1 - √15 i)/2.
Does x² + x + 4 = 0 factor?
Not over the integers, because the discriminant -15 is negative. Use the quadratic formula.
What is the vertex of y = x² + x + 4?
(-1/2, 15/4). 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 = 4.