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Solve 3x² + 8x + 11 = 0

3x² + 8x + 11 = 0 has no real solutions. The two complex solutions are x = (-4 + √17 i)/3 and x = (-4 - √17 i)/3.

x² +x += 0
3x2+8x+11=0⇒x=−4+17 i3, −4−17 i3 (complex)3x^2 + 8x + 11 = 0 \Rightarrow x = \frac{-4 + \sqrt{17}\, i}{3},\ \frac{-4 - \sqrt{17}\, i}{3}\ \text{(complex)}
Discriminant D = -68

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

Graph of y = 3x² + 8x + 11-5-4-3-2-1121020304050vertexy = 3x² + 8x + 11
Graph of y = 3x² + 8x + 11. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.3333333, 5.6666667).

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 = 8, c = 11.

    3x2+8x+11=03x^2 + 8x + 11 = 0
  2. 2

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

    D=82−4⋅3⋅11=64−132=−68D = 8^2 - 4\cdot3\cdot11 = 64 - 132 = -68
  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=−8±−686x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{-68}}{6}
  5. 5

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

    x=−4+17 i3orx=−4−17 i3x = \frac{-4 + \sqrt{17}\, i}{3}\quad\text{or}\quad x = \frac{-4 - \sqrt{17}\, i}{3}
LaTeX source of the formulas on this page
\[ 3x^2 + 8x + 11 = 0\ \Rightarrow\ x = \frac{-4 + \sqrt{17}\, i}{3},\ \frac{-4 - \sqrt{17}\, i}{3} \]
\[ 3x^2 + 8x + 11 = 0 \]
\[ D = 8^2 - 4\cdot3\cdot11 = 64 - 132 = -68 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{-68}}{6} \]
\[ x = \frac{-4 + \sqrt{17}\, i}{3}\quad\text{or}\quad x = \frac{-4 - \sqrt{17}\, i}{3} \]

The parabola y = 3x² + 8x + 11

xy
-4.333332.666067
-3.333317.666267
-2.33338.6664667
-1.33335.6666667
-0.33338.6668667
0.666717.667067
1.666732.667267

Questions

What are the solutions of 3x² + 8x + 11 = 0?

3x² + 8x + 11 = 0 has no real solutions. The two complex solutions are x = (-4 + √17 i)/3 and x = (-4 - √17 i)/3.

Does 3x² + 8x + 11 = 0 factor?

Not over the integers, because the discriminant -68 is negative. Use the quadratic formula.

What is the vertex of y = 3x² + 8x + 11?

(-4/3, 17/3). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -8/3, product = c/a = 11/3.

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