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Solve 3x² + 11x + 11 = 0

3x² + 11x + 11 = 0 has no real solutions. The two complex solutions are x = (-11 + √11 i)/6 and x = (-11 - √11 i)/6.

x² +x += 0
3x2+11x+11=0⇒x=−11+11 i6, −11−11 i6 (complex)3x^2 + 11x + 11 = 0 \Rightarrow x = \frac{-11 + \sqrt{11}\, i}{6},\ \frac{-11 - \sqrt{11}\, i}{6}\ \text{(complex)}
Discriminant D = -11

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

Graph of y = 3x² + 11x + 11-5-4-3-2-1121020304050vertexy = 3x² + 11x + 11
Graph of y = 3x² + 11x + 11. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.8333333, 0.91666667).

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 = 11, c = 11.

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

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

    D=112−4⋅3⋅11=121−132=−11D = 11^2 - 4\cdot3\cdot11 = 121 - 132 = -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=−11±−116x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{-11}}{6}
  5. 5

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

    x=−11+11 i6orx=−11−11 i6x = \frac{-11 + \sqrt{11}\, i}{6}\quad\text{or}\quad x = \frac{-11 - \sqrt{11}\, i}{6}
LaTeX source of the formulas on this page
\[ 3x^2 + 11x + 11 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{11}\, i}{6},\ \frac{-11 - \sqrt{11}\, i}{6} \]
\[ 3x^2 + 11x + 11 = 0 \]
\[ D = 11^2 - 4\cdot3\cdot11 = 121 - 132 = -11 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{-11}}{6} \]
\[ x = \frac{-11 + \sqrt{11}\, i}{6}\quad\text{or}\quad x = \frac{-11 - \sqrt{11}\, i}{6} \]

The parabola y = 3x² + 11x + 11

xy
-4.833327.916067
-3.833312.916267
-2.83333.9164667
-1.83330.91666667
-0.83333.9168667
0.166712.917067
1.166727.917267

Questions

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

3x² + 11x + 11 = 0 has no real solutions. The two complex solutions are x = (-11 + √11 i)/6 and x = (-11 - √11 i)/6.

Does 3x² + 11x + 11 = 0 factor?

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

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

(-11/6, 11/12). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

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

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