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Solve -3x² + 2x - 1 = 0

-3x² + 2x - 1 = 0 has no real solutions. The two complex solutions are x = (1 + √2 i)/3 and x = (1 - √2 i)/3.

x² +x += 0
−3x2+2x−1=0⇒x=1+2 i3, 1−2 i3 (complex)-3x^2 + 2x - 1 = 0 \Rightarrow x = \frac{1 + \sqrt{2}\, i}{3},\ \frac{1 - \sqrt{2}\, i}{3}\ \text{(complex)}
Discriminant D = -8

Change a, b or c (whole numbers, a not 0) and the exact solutions follow.

Graph of y = -3x² + 2x - 1-3-2-11234-50-40-30-20-10vertexy = -3x² + 2x - 1
Graph of y = -3x² + 2x - 1. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.33333333, -0.66666667).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -3, b = 2, c = -1.

    −3x2+2x−1=0-3x^2 + 2x - 1 = 0
  2. 2

    Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.

    D=22−4⋅(−3)⋅(−1)=4−12=−8D = 2^2 - 4\cdot(-3)\cdot(-1) = 4 - 12 = -8
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−2±−8−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-8}}{-6}
  5. 5

    Simplify (pull the square out of the root) to get the solutions:

    x=1+2 i3orx=1−2 i3x = \frac{1 + \sqrt{2}\, i}{3}\quad\text{or}\quad x = \frac{1 - \sqrt{2}\, i}{3}
LaTeX source of the formulas on this page
\[ -3x^2 + 2x - 1 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{2}\, i}{3},\ \frac{1 - \sqrt{2}\, i}{3} \]
\[ -3x^2 + 2x - 1 = 0 \]
\[ D = 2^2 - 4\cdot(-3)\cdot(-1) = 4 - 12 = -8 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-8}}{-6} \]
\[ x = \frac{1 + \sqrt{2}\, i}{3}\quad\text{or}\quad x = \frac{1 - \sqrt{2}\, i}{3} \]

The parabola y = -3x² + 2x - 1

xy
-2.6667-27.667267
-1.6667-12.667067
-0.6667-3.6668667
0.3333-0.66666667
1.3333-3.6664667
2.3333-12.666267
3.3333-27.666067

Questions

What are the solutions of -3x² + 2x - 1 = 0?

-3x² + 2x - 1 = 0 has no real solutions. The two complex solutions are x = (1 + √2 i)/3 and x = (1 - √2 i)/3.

Does -3x² + 2x - 1 = 0 factor?

Not over the integers, because the discriminant -8 is negative. Use the quadratic formula.

What is the vertex of y = -3x² + 2x - 1?

(1/3, -2/3). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 2/3, product = c/a = 1/3.

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