Solve 3x² + 6x + 1 = 0
The solutions of 3x² + 6x + 1 = 0 are x = (-3 + √6)/3 or x = (-3 - √6)/3, that is x ≈ -0.18350342 and x ≈ -1.8164966.
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 = 3, b = 6, c = 1.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is positive, so there are two different real solutions.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ -0.18350342 and x ≈ -1.8164966.
- 7
Check numerically: for x ≈ -0.18350342 the left side is -4.4408921e-16 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 3x^2 + 6x + 1 = 0\ \Rightarrow\ x = \frac{-3 + \sqrt{6}}{3},\ \frac{-3 - \sqrt{6}}{3} \]
\[ 3x^2 + 6x + 1 = 0 \]
\[ D = 6^2 - 4\cdot3\cdot1 = 36 - 12 = 24 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{24}}{6} \]
\[ x = \frac{-3 + \sqrt{6}}{3}\quad\text{or}\quad x = \frac{-3 - \sqrt{6}}{3} \]The parabola y = 3x² + 6x + 1
| x | y |
|---|---|
| -4 | 25 |
| -3 | 10 |
| -2 | 1 |
| -1 | -2 |
| 0 | 1 |
| 1 | 10 |
| 2 | 25 |
Questions
What are the solutions of 3x² + 6x + 1 = 0?
The solutions of 3x² + 6x + 1 = 0 are x = (-3 + √6)/3 or x = (-3 - √6)/3, that is x ≈ -0.18350342 and x ≈ -1.8164966.
Does 3x² + 6x + 1 = 0 factor?
Not over the integers, because the discriminant 24 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 3x² + 6x + 1?
(-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 = 1/3.