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Solve -3x² - 3 = 0

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

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

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

Graph of y = -3x² - 3-4-3-2-11234-50-40-30-20-10vertexy = -3x² - 3
Graph of y = -3x² - 3. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, -3).

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 = 0, c = -3.

    −3x2−3=0-3x^2 - 3 = 0
  2. 2

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

    D=02−4⋅(−3)⋅(−3)=0−36=−36D = 0^2 - 4\cdot(-3)\cdot(-3) = 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±−36−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-36}}{-6}
  5. 5

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

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

The parabola y = -3x² - 3

xy
-3-30
-2-15
-1-6
0-3
1-6
2-15
3-30

Questions

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

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

Does -3x² - 3 = 0 factor?

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

What is the vertex of y = -3x² - 3?

(0, -3). 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 = 1.

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