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Solve 3x² + 3x + 9 = 0

3x² + 3x + 9 = 0 has no real solutions. The two complex solutions are x = (-1 + √11 i)/2 and x = (-1 - √11 i)/2.

x² +x += 0
3x2+3x+9=0⇒x=−1+11 i2, −1−11 i2 (complex)3x^2 + 3x + 9 = 0 \Rightarrow x = \frac{-1 + \sqrt{11}\, i}{2},\ \frac{-1 - \sqrt{11}\, i}{2}\ \text{(complex)}
Discriminant D = -99

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

Graph of y = 3x² + 3x + 9-4-3-2-1123102030405060vertexy = 3x² + 3x + 9
Graph of y = 3x² + 3x + 9. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-0.5, 8.25).

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 = 3, c = 9.

    3x2+3x+9=03x^2 + 3x + 9 = 0
  2. 2

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

    D=32−4⋅3⋅9=9−108=−99D = 3^2 - 4\cdot3\cdot9 = 9 - 108 = -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=−3±−996x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-99}}{6}
  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
\[ 3x^2 + 3x + 9 = 0\ \Rightarrow\ x = \frac{-1 + \sqrt{11}\, i}{2},\ \frac{-1 - \sqrt{11}\, i}{2} \]
\[ 3x^2 + 3x + 9 = 0 \]
\[ D = 3^2 - 4\cdot3\cdot9 = 9 - 108 = -99 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-99}}{6} \]
\[ x = \frac{-1 + \sqrt{11}\, i}{2}\quad\text{or}\quad x = \frac{-1 - \sqrt{11}\, i}{2} \]

The parabola y = 3x² + 3x + 9

xy
-3.535.25
-2.520.25
-1.511.25
-0.58.25
0.511.25
1.520.25
2.535.25

Questions

What are the solutions of 3x² + 3x + 9 = 0?

3x² + 3x + 9 = 0 has no real solutions. The two complex solutions are x = (-1 + √11 i)/2 and x = (-1 - √11 i)/2.

Does 3x² + 3x + 9 = 0 factor?

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

What is the vertex of y = 3x² + 3x + 9?

(-1/2, 33/4). 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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