Solve 4x² + 8x + 6 = 0
4x² + 8x + 6 = 0 has no real solutions. The two complex solutions are x = (-2 + √2 i)/2 and x = (-2 - √2 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 = 4, b = 8, c = 6.
- 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
\[ 4x^2 + 8x + 6 = 0\ \Rightarrow\ x = \frac{-2 + \sqrt{2}\, i}{2},\ \frac{-2 - \sqrt{2}\, i}{2} \]
\[ 4x^2 + 8x + 6 = 0 \]
\[ D = 8^2 - 4\cdot4\cdot6 = 64 - 96 = -32 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{-32}}{8} \]
\[ x = \frac{-2 + \sqrt{2}\, i}{2}\quad\text{or}\quad x = \frac{-2 - \sqrt{2}\, i}{2} \]The parabola y = 4x² + 8x + 6
| x | y |
|---|---|
| -4 | 38 |
| -3 | 18 |
| -2 | 6 |
| -1 | 2 |
| 0 | 6 |
| 1 | 18 |
| 2 | 38 |
Questions
What are the solutions of 4x² + 8x + 6 = 0?
4x² + 8x + 6 = 0 has no real solutions. The two complex solutions are x = (-2 + √2 i)/2 and x = (-2 - √2 i)/2.
Does 4x² + 8x + 6 = 0 factor?
Not over the integers, because the discriminant -32 is negative. Use the quadratic formula.
What is the vertex of y = 4x² + 8x + 6?
(-1, 2). 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 = 3/2.