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Solve -x² - 6x - 11 = 0

-x² - 6x - 11 = 0 has no real solutions. The two complex solutions are x = -3 + √2 i and x = -3 - √2 i.

x² +x += 0
−x2−6x−11=0⇒x=−3+2 i, −3−2 i (complex)-x^2 - 6x - 11 = 0 \Rightarrow x = -3 + \sqrt{2}\, i,\ -3 - \sqrt{2}\, i\ \text{(complex)}
Discriminant D = -8

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

Graph of y = -x² - 6x - 11-7-6-5-4-3-2-11-18-16-14-12-10-8-6-4-2vertexy = -x² - 6x - 11
Graph of y = -x² - 6x - 11. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-3, -2).

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 = -6, c = -11.

    −x2−6x−11=0-x^2 - 6x - 11 = 0
  2. 2

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

    D=(−6)2−4⋅(−1)⋅(−11)=36−44=−8D = (-6)^2 - 4\cdot(-1)\cdot(-11) = 36 - 44 = -8
  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=6±−8−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{-8}}{-2}
  5. 5

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

    x=−3+2 iorx=−3−2 ix = -3 + \sqrt{2}\, i\quad\text{or}\quad x = -3 - \sqrt{2}\, i
LaTeX source of the formulas on this page
\[ -x^2 - 6x - 11 = 0\ \Rightarrow\ x = -3 + \sqrt{2}\, i,\ -3 - \sqrt{2}\, i \]
\[ -x^2 - 6x - 11 = 0 \]
\[ D = (-6)^2 - 4\cdot(-1)\cdot(-11) = 36 - 44 = -8 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{-8}}{-2} \]
\[ x = -3 + \sqrt{2}\, i\quad\text{or}\quad x = -3 - \sqrt{2}\, i \]

The parabola y = -x² - 6x - 11

xy
-6-11
-5-6
-4-3
-3-2
-2-3
-1-6
0-11

Questions

What are the solutions of -x² - 6x - 11 = 0?

-x² - 6x - 11 = 0 has no real solutions. The two complex solutions are x = -3 + √2 i and x = -3 - √2 i.

Does -x² - 6x - 11 = 0 factor?

Not over the integers, because the discriminant -8 is negative. Use the quadratic formula.

What is the vertex of y = -x² - 6x - 11?

(-3, -2). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = -6, product = c/a = 11.

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