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Solve x² + 5x + 11 = 0

x² + 5x + 11 = 0 has no real solutions. The two complex solutions are x = (-5 + √19 i)/2 and x = (-5 - √19 i)/2.

x² +x += 0
x2+5x+11=0⇒x=−5+19 i2, −5−19 i2 (complex)x^2 + 5x + 11 = 0 \Rightarrow x = \frac{-5 + \sqrt{19}\, i}{2},\ \frac{-5 - \sqrt{19}\, i}{2}\ \text{(complex)}
Discriminant D = -19

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

Graph of y = x² + 5x + 11-6-5-4-3-2-1146810121416182022vertexy = x² + 5x + 11
Graph of y = x² + 5x + 11. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-2.5, 4.75).

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 = 5, c = 11.

    x2+5x+11=0x^2 + 5x + 11 = 0
  2. 2

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

    D=52−4⋅1⋅11=25−44=−19D = 5^2 - 4\cdot1\cdot11 = 25 - 44 = -19
  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=−5±−192x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-19}}{2}
  5. 5

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

    x=−5+19 i2orx=−5−19 i2x = \frac{-5 + \sqrt{19}\, i}{2}\quad\text{or}\quad x = \frac{-5 - \sqrt{19}\, i}{2}
LaTeX source of the formulas on this page
\[ x^2 + 5x + 11 = 0\ \Rightarrow\ x = \frac{-5 + \sqrt{19}\, i}{2},\ \frac{-5 - \sqrt{19}\, i}{2} \]
\[ x^2 + 5x + 11 = 0 \]
\[ D = 5^2 - 4\cdot1\cdot11 = 25 - 44 = -19 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-19}}{2} \]
\[ x = \frac{-5 + \sqrt{19}\, i}{2}\quad\text{or}\quad x = \frac{-5 - \sqrt{19}\, i}{2} \]

The parabola y = x² + 5x + 11

xy
-5.513.75
-4.58.75
-3.55.75
-2.54.75
-1.55.75
-0.58.75
0.513.75

Questions

What are the solutions of x² + 5x + 11 = 0?

x² + 5x + 11 = 0 has no real solutions. The two complex solutions are x = (-5 + √19 i)/2 and x = (-5 - √19 i)/2.

Does x² + 5x + 11 = 0 factor?

Not over the integers, because the discriminant -19 is negative. Use the quadratic formula.

What is the vertex of y = x² + 5x + 11?

(-5/2, 19/4). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -5, product = c/a = 11.

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