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Solve -3x² - x - 11 = 0

-3x² - x - 11 = 0 has no real solutions. The two complex solutions are x = (-1 + √131 i)/6 and x = (-1 - √131 i)/6.

x² +x += 0
−3x2−x−11=0⇒x=−1+131 i6, −1−131 i6 (complex)-3x^2 - x - 11 = 0 \Rightarrow x = \frac{-1 + \sqrt{131}\, i}{6},\ \frac{-1 - \sqrt{131}\, i}{6}\ \text{(complex)}
Discriminant D = -131

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

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Graph of y = -3x² - x - 11-4-3-2-1123-60-50-40-30-20-10vertexy = -3x² - x - 11
Graph of y = -3x² - x - 11. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-0.16666667, -10.916667).

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 = -1, c = -11.

    −3x2−x−11=0-3x^2 - x - 11 = 0
  2. 2

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

    D=(−1)2−4⋅(−3)⋅(−11)=1−132=−131D = (-1)^2 - 4\cdot(-3)\cdot(-11) = 1 - 132 = -131
  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±−131−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-131}}{-6}
  5. 5

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

    x=−1+131 i6orx=−1−131 i6x = \frac{-1 + \sqrt{131}\, i}{6}\quad\text{or}\quad x = \frac{-1 - \sqrt{131}\, i}{6}
LaTeX source of the formulas on this page
\[ -3x^2 - x - 11 = 0\ \Rightarrow\ x = \frac{-1 + \sqrt{131}\, i}{6},\ \frac{-1 - \sqrt{131}\, i}{6} \]
\[ -3x^2 - x - 11 = 0 \]
\[ D = (-1)^2 - 4\cdot(-3)\cdot(-11) = 1 - 132 = -131 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{-131}}{-6} \]
\[ x = \frac{-1 + \sqrt{131}\, i}{6}\quad\text{or}\quad x = \frac{-1 - \sqrt{131}\, i}{6} \]

The parabola y = -3x² - x - 11

xy
-3.1667-37.917267
-2.1667-22.917067
-1.1667-13.916867
-0.1667-10.916667
0.8333-13.916467
1.8333-22.916267
2.8333-37.916067

Questions

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

-3x² - x - 11 = 0 has no real solutions. The two complex solutions are x = (-1 + √131 i)/6 and x = (-1 - √131 i)/6.

Does -3x² - x - 11 = 0 factor?

Not over the integers, because the discriminant -131 is negative. Use the quadratic formula.

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

(-1/6, -131/12). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = -1/3, product = c/a = 11/3.

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