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Solve -x² + 3x - 3 = 0

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

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

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

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Graph of y = -x² + 3x - 3-2-112345-16-14-12-10-8-6-4-2vertexy = -x² + 3x - 3
Graph of y = -x² + 3x - 3. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (1.5, -0.75).

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 = 3, c = -3.

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

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

    D=32−4⋅(−1)⋅(−3)=9−12=−3D = 3^2 - 4\cdot(-1)\cdot(-3) = 9 - 12 = -3
  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±−3−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-3}}{-2}
  5. 5

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

    x=3+3 i2orx=3−3 i2x = \frac{3 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{3}\, i}{2}
LaTeX source of the formulas on this page
\[ -x^2 + 3x - 3 = 0\ \Rightarrow\ x = \frac{3 + \sqrt{3}\, i}{2},\ \frac{3 - \sqrt{3}\, i}{2} \]
\[ -x^2 + 3x - 3 = 0 \]
\[ D = 3^2 - 4\cdot(-1)\cdot(-3) = 9 - 12 = -3 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{-3}}{-2} \]
\[ x = \frac{3 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{3 - \sqrt{3}\, i}{2} \]

The parabola y = -x² + 3x - 3

xy
-1.5-9.75
-0.5-4.75
0.5-1.75
1.5-0.75
2.5-1.75
3.5-4.75
4.5-9.75

Questions

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

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

Does -x² + 3x - 3 = 0 factor?

Not over the integers, because the discriminant -3 is negative. Use the quadratic formula.

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

(3/2, -3/4). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 3, product = c/a = 3.

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