Solve -3x² + x - 1 = 0
-3x² + x - 1 = 0 has no real solutions. The two complex solutions are x = (1 + √11 i)/6 and x = (1 - √11 i)/6.
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 = 1, 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
\[ -3x^2 + x - 1 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{11}\, i}{6},\ \frac{1 - \sqrt{11}\, i}{6} \]
\[ -3x^2 + x - 1 = 0 \]
\[ D = 1^2 - 4\cdot(-3)\cdot(-1) = 1 - 12 = -11 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{-11}}{-6} \]
\[ x = \frac{1 + \sqrt{11}\, i}{6}\quad\text{or}\quad x = \frac{1 - \sqrt{11}\, i}{6} \]The parabola y = -3x² + x - 1
| x | y |
|---|---|
| -2.8333 | -27.916067 |
| -1.8333 | -12.916267 |
| -0.8333 | -3.9164667 |
| 0.1667 | -0.91666667 |
| 1.1667 | -3.9168667 |
| 2.1667 | -12.917067 |
| 3.1667 | -27.917267 |
Questions
What are the solutions of -3x² + x - 1 = 0?
-3x² + x - 1 = 0 has no real solutions. The two complex solutions are x = (1 + √11 i)/6 and x = (1 - √11 i)/6.
Does -3x² + x - 1 = 0 factor?
Not over the integers, because the discriminant -11 is negative. Use the quadratic formula.
What is the vertex of y = -3x² + x - 1?
(1/6, -11/12). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = 1/3, product = c/a = 1/3.