Solve 3x² - 12x - 1 = 0
The solutions of 3x² - 12x - 1 = 0 are x = (6 + √39)/3 or x = (6 - √39)/3, that is x ≈ 4.081666 and x ≈ -0.081665999.
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 = -12, 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 ≈ 4.081666 and x ≈ -0.081665999.
- 7
Check numerically: for x ≈ 4.081666 the left side is 7.1054274e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ 3x^2 - 12x - 1 = 0\ \Rightarrow\ x = \frac{6 + \sqrt{39}}{3},\ \frac{6 - \sqrt{39}}{3} \]
\[ 3x^2 - 12x - 1 = 0 \]
\[ D = (-12)^2 - 4\cdot3\cdot(-1) = 144 + 12 = 156 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{156}}{6} \]
\[ x = \frac{6 + \sqrt{39}}{3}\quad\text{or}\quad x = \frac{6 - \sqrt{39}}{3} \]The parabola y = 3x² - 12x - 1
| x | y |
|---|---|
| -1 | 14 |
| 0 | -1 |
| 1 | -10 |
| 2 | -13 |
| 3 | -10 |
| 4 | -1 |
| 5 | 14 |
Questions
What are the solutions of 3x² - 12x - 1 = 0?
The solutions of 3x² - 12x - 1 = 0 are x = (6 + √39)/3 or x = (6 - √39)/3, that is x ≈ 4.081666 and x ≈ -0.081665999.
Does 3x² - 12x - 1 = 0 factor?
Not over the integers, because the discriminant 156 is not a perfect square. Use the quadratic formula.
What is the vertex of y = 3x² - 12x - 1?
(2, -13). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = 4, product = c/a = -1/3.