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Solve x² - 2x + 2 = 0

x² - 2x + 2 = 0 has no real solutions. The two complex solutions are x = 1 + i and x = 1 - i.

x² +x += 0
x2−2x+2=0⇒x=1+i, 1−i (complex)x^2 - 2x + 2 = 0 \Rightarrow x = 1 + i,\ 1 - i\ \text{(complex)}
Discriminant D = -4

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

Graph of y = x² - 2x + 2-3-2-11234524681012141618vertexy = x² - 2x + 2
Graph of y = x² - 2x + 2. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (1, 1).

How to work it out

  1. 1

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

    x2−2x+2=0x^2 - 2x + 2 = 0
  2. 2

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

    D=(−2)2−4⋅1⋅2=4−8=−4D = (-2)^2 - 4\cdot1\cdot2 = 4 - 8 = -4
  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±−42x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm \sqrt{-4}}{2}
  5. 5

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

    x=1+iorx=1−ix = 1 + i\quad\text{or}\quad x = 1 - i
LaTeX source of the formulas on this page
\[ x^2 - 2x + 2 = 0\ \Rightarrow\ x = 1 + i,\ 1 - i \]
\[ x^2 - 2x + 2 = 0 \]
\[ D = (-2)^2 - 4\cdot1\cdot2 = 4 - 8 = -4 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm \sqrt{-4}}{2} \]
\[ x = 1 + i\quad\text{or}\quad x = 1 - i \]

The parabola y = x² - 2x + 2

xy
-210
-15
02
11
22
35
410

Questions

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

x² - 2x + 2 = 0 has no real solutions. The two complex solutions are x = 1 + i and x = 1 - i.

Does x² - 2x + 2 = 0 factor?

Not over the integers, because the discriminant -4 is negative. Use the quadratic formula.

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

(1, 1). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 2, product = c/a = 2.

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