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Solve x² - x + 3 = 0

x² - x + 3 = 0 has no real solutions. The two complex solutions are x = (1 + √11 i)/2 and x = (1 - √11 i)/2.

x² +x += 0
x2−x+3=0⇒x=1+11 i2, 1−11 i2 (complex)x^2 - x + 3 = 0 \Rightarrow x = \frac{1 + \sqrt{11}\, i}{2},\ \frac{1 - \sqrt{11}\, i}{2}\ \text{(complex)}
Discriminant D = -11

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

Graph of y = x² - x + 3-3-2-112342468101214161820vertexy = x² - x + 3
Graph of y = x² - x + 3. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.5, 2.75).

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 = -1, c = 3.

    x2−x+3=0x^2 - x + 3 = 0
  2. 2

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

    D=(−1)2−4⋅1⋅3=1−12=−11D = (-1)^2 - 4\cdot1\cdot3 = 1 - 12 = -11
  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=1±−112x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-11}}{2}
  5. 5

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

    x=1+11 i2orx=1−11 i2x = \frac{1 + \sqrt{11}\, i}{2}\quad\text{or}\quad x = \frac{1 - \sqrt{11}\, i}{2}
LaTeX source of the formulas on this page
\[ x^2 - x + 3 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{11}\, i}{2},\ \frac{1 - \sqrt{11}\, i}{2} \]
\[ x^2 - x + 3 = 0 \]
\[ D = (-1)^2 - 4\cdot1\cdot3 = 1 - 12 = -11 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-11}}{2} \]
\[ x = \frac{1 + \sqrt{11}\, i}{2}\quad\text{or}\quad x = \frac{1 - \sqrt{11}\, i}{2} \]

The parabola y = x² - x + 3

xy
-2.511.75
-1.56.75
-0.53.75
0.52.75
1.53.75
2.56.75
3.511.75

Questions

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

x² - x + 3 = 0 has no real solutions. The two complex solutions are x = (1 + √11 i)/2 and x = (1 - √11 i)/2.

Does x² - x + 3 = 0 factor?

Not over the integers, because the discriminant -11 is negative. Use the quadratic formula.

What is the vertex of y = x² - x + 3?

(1/2, 11/4). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 1, product = c/a = 3.

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