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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 + 1-3-2-112341020304050vertexy = 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=03x^2 - 2x + 1 = 0
  2. 2

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

    D=(−2)2−4⋅3⋅1=4−12=−8D = (-2)^2 - 4\cdot3\cdot1 = 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±−86x = \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
\[ 3x^2 - 2x + 1 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{2}\, i}{3},\ \frac{1 - \sqrt{2}\, i}{3} \]
\[ 3x^2 - 2x + 1 = 0 \]
\[ D = (-2)^2 - 4\cdot3\cdot1 = 4 - 12 = -8 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm \sqrt{-8}}{6} \]
\[ x = \frac{1 + \sqrt{2}\, i}{3}\quad\text{or}\quad x = \frac{1 - \sqrt{2}\, i}{3} \]

The parabola y = 3x² - 2x + 1

xy
-2.666727.667267
-1.666712.667067
-0.66673.6668667
0.33330.66666667
1.33333.6664667
2.333312.666267
3.333327.666067

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 upwards, so this is the minimum.

What are the sum and product of the roots?

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

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