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Solve 5x² - x + 5 = 0

5x² - x + 5 = 0 has no real solutions. The two complex solutions are x = (1 + 3√11 i)/10 and x = (1 - 3√11 i)/10.

x² +x += 0
5x2−x+5=0⇒x=1+311 i10, 1−311 i10 (complex)5x^2 - x + 5 = 0 \Rightarrow x = \frac{1 + 3\sqrt{11}\, i}{10},\ \frac{1 - 3\sqrt{11}\, i}{10}\ \text{(complex)}
Discriminant D = -99

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

Graph of y = 5x² - x + 5-3-2-1123420406080vertexy = 5x² - x + 5
Graph of y = 5x² - x + 5. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.1, 4.95).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 5, b = -1, c = 5.

    5x2−x+5=05x^2 - x + 5 = 0
  2. 2

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

    D=(−1)2−4⋅5⋅5=1−100=−99D = (-1)^2 - 4\cdot5\cdot5 = 1 - 100 = -99
  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±−9910x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-99}}{10}
  5. 5

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

    x=1+311 i10orx=1−311 i10x = \frac{1 + 3\sqrt{11}\, i}{10}\quad\text{or}\quad x = \frac{1 - 3\sqrt{11}\, i}{10}
LaTeX source of the formulas on this page
\[ 5x^2 - x + 5 = 0\ \Rightarrow\ x = \frac{1 + 3\sqrt{11}\, i}{10},\ \frac{1 - 3\sqrt{11}\, i}{10} \]
\[ 5x^2 - x + 5 = 0 \]
\[ D = (-1)^2 - 4\cdot5\cdot5 = 1 - 100 = -99 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-99}}{10} \]
\[ x = \frac{1 + 3\sqrt{11}\, i}{10}\quad\text{or}\quad x = \frac{1 - 3\sqrt{11}\, i}{10} \]

The parabola y = 5x² - x + 5

xy
-2.949.95
-1.924.95
-0.99.95
0.14.95
1.19.95
2.124.95
3.149.95

Questions

What are the solutions of 5x² - x + 5 = 0?

5x² - x + 5 = 0 has no real solutions. The two complex solutions are x = (1 + 3√11 i)/10 and x = (1 - 3√11 i)/10.

Does 5x² - x + 5 = 0 factor?

Not over the integers, because the discriminant -99 is negative. Use the quadratic formula.

What is the vertex of y = 5x² - x + 5?

(1/10, 99/20). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 1/5, product = c/a = 1.

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