Solve 5x² + 4x + 1 = 0
5x² + 4x + 1 = 0 has no real solutions. The two complex solutions are x = (-2 + i)/5 and x = (-2 - i)/5.
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 = 5, b = 4, 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
\[ 5x^2 + 4x + 1 = 0\ \Rightarrow\ x = \frac{-2 + i}{5},\ \frac{-2 - i}{5} \]
\[ 5x^2 + 4x + 1 = 0 \]
\[ D = 4^2 - 4\cdot5\cdot1 = 16 - 20 = -4 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-4 \pm \sqrt{-4}}{10} \]
\[ x = \frac{-2 + i}{5}\quad\text{or}\quad x = \frac{-2 - i}{5} \]The parabola y = 5x² + 4x + 1
| x | y |
|---|---|
| -3.4 | 45.2 |
| -2.4 | 20.2 |
| -1.4 | 5.2 |
| -0.4 | 0.2 |
| 0.6 | 5.2 |
| 1.6 | 20.2 |
| 2.6 | 45.2 |
Questions
What are the solutions of 5x² + 4x + 1 = 0?
5x² + 4x + 1 = 0 has no real solutions. The two complex solutions are x = (-2 + i)/5 and x = (-2 - i)/5.
Does 5x² + 4x + 1 = 0 factor?
Not over the integers, because the discriminant -4 is negative. Use the quadratic formula.
What is the vertex of y = 5x² + 4x + 1?
(-2/5, 1/5). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -4/5, product = c/a = 1/5.