Solve -3x² + 6x + 1 = 0
The solutions of -3x² + 6x + 1 = 0 are x = (3 + 2√3)/3 or x = (3 - 2√3)/3, that is x ≈ -0.15470054 and x ≈ 2.1547005.
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 ≈ -0.15470054 and x ≈ 2.1547005.
- 7
Check numerically: for x ≈ -0.15470054 the left side is 4.4408921e-16 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ -3x^2 + 6x + 1 = 0\ \Rightarrow\ x = \frac{3 + 2\sqrt{3}}{3},\ \frac{3 - 2\sqrt{3}}{3} \]
\[ -3x^2 + 6x + 1 = 0 \]
\[ D = 6^2 - 4\cdot(-3)\cdot1 = 36 + 12 = 48 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{48}}{-6} \]
\[ x = \frac{3 + 2\sqrt{3}}{3}\quad\text{or}\quad x = \frac{3 - 2\sqrt{3}}{3} \]The parabola y = -3x² + 6x + 1
| x | y |
|---|---|
| -2 | -23 |
| -1 | -8 |
| 0 | 1 |
| 1 | 4 |
| 2 | 1 |
| 3 | -8 |
| 4 | -23 |
Questions
What are the solutions of -3x² + 6x + 1 = 0?
The solutions of -3x² + 6x + 1 = 0 are x = (3 + 2√3)/3 or x = (3 - 2√3)/3, that is x ≈ -0.15470054 and x ≈ 2.1547005.
Does -3x² + 6x + 1 = 0 factor?
Not over the integers, because the discriminant 48 is not a perfect square. Use the quadratic formula.
What is the vertex of y = -3x² + 6x + 1?
(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 = -1/3.