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Solve 3x² - 11x + 12 = 0

3x² - 11x + 12 = 0 has no real solutions. The two complex solutions are x = (11 + √23 i)/6 and x = (11 - √23 i)/6.

x² +x += 0
3x2−11x+12=0⇒x=11+23 i6, 11−23 i6 (complex)3x^2 - 11x + 12 = 0 \Rightarrow x = \frac{11 + \sqrt{23}\, i}{6},\ \frac{11 - \sqrt{23}\, i}{6}\ \text{(complex)}
Discriminant D = -23

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

Graph of y = 3x² - 11x + 12-2-1123451020304050vertexy = 3x² - 11x + 12
Graph of y = 3x² - 11x + 12. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (1.8333333, 1.9166667).

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 = -11, c = 12.

    3x2−11x+12=03x^2 - 11x + 12 = 0
  2. 2

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

    D=(−11)2−4⋅3⋅12=121−144=−23D = (-11)^2 - 4\cdot3\cdot12 = 121 - 144 = -23
  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=11±−236x = \frac{-b \pm \sqrt{D}}{2a} = \frac{11 \pm \sqrt{-23}}{6}
  5. 5

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

    x=11+23 i6orx=11−23 i6x = \frac{11 + \sqrt{23}\, i}{6}\quad\text{or}\quad x = \frac{11 - \sqrt{23}\, i}{6}
LaTeX source of the formulas on this page
\[ 3x^2 - 11x + 12 = 0\ \Rightarrow\ x = \frac{11 + \sqrt{23}\, i}{6},\ \frac{11 - \sqrt{23}\, i}{6} \]
\[ 3x^2 - 11x + 12 = 0 \]
\[ D = (-11)^2 - 4\cdot3\cdot12 = 121 - 144 = -23 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{11 \pm \sqrt{-23}}{6} \]
\[ x = \frac{11 + \sqrt{23}\, i}{6}\quad\text{or}\quad x = \frac{11 - \sqrt{23}\, i}{6} \]

The parabola y = 3x² - 11x + 12

xy
-1.166728.917267
-0.166713.917067
0.83334.9168667
1.83331.9166667
2.83334.9164667
3.833313.916267
4.833328.916067

Questions

What are the solutions of 3x² - 11x + 12 = 0?

3x² - 11x + 12 = 0 has no real solutions. The two complex solutions are x = (11 + √23 i)/6 and x = (11 - √23 i)/6.

Does 3x² - 11x + 12 = 0 factor?

Not over the integers, because the discriminant -23 is negative. Use the quadratic formula.

What is the vertex of y = 3x² - 11x + 12?

(11/6, 23/12). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 11/3, product = c/a = 4.

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