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Solve -x² - 9 = 0

-x² - 9 = 0 has no real solutions. The two complex solutions are x = 3i and x = -3i.

x² +x += 0
−x2−9=0⇒x=3i, −3i (complex)-x^2 - 9 = 0 \Rightarrow x = 3i,\ -3i\ \text{(complex)}
Discriminant D = -36

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

Graph of y = -x² - 9-4-3-2-11234-26-24-22-20-18-16-14-12-10-8vertexy = -x² - 9
Graph of y = -x² - 9. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, -9).

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 = 0, c = -9.

    −x2−9=0-x^2 - 9 = 0
  2. 2

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

    D=02−4⋅(−1)⋅(−9)=0−36=−36D = 0^2 - 4\cdot(-1)\cdot(-9) = 0 - 36 = -36
  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=0±−36−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-36}}{-2}
  5. 5

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

    x=3iorx=−3ix = 3i\quad\text{or}\quad x = -3i
LaTeX source of the formulas on this page
\[ -x^2 - 9 = 0\ \Rightarrow\ x = 3i,\ -3i \]
\[ -x^2 - 9 = 0 \]
\[ D = 0^2 - 4\cdot(-1)\cdot(-9) = 0 - 36 = -36 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-36}}{-2} \]
\[ x = 3i\quad\text{or}\quad x = -3i \]

The parabola y = -x² - 9

xy
-3-18
-2-13
-1-10
0-9
1-10
2-13
3-18

Questions

What are the solutions of -x² - 9 = 0?

-x² - 9 = 0 has no real solutions. The two complex solutions are x = 3i and x = -3i.

Does -x² - 9 = 0 factor?

Not over the integers, because the discriminant -36 is negative. Use the quadratic formula.

What is the vertex of y = -x² - 9?

(0, -9). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 0, product = c/a = 9.

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