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Solve -3x² - x + 1 = 0

The solutions of -3x² - x + 1 = 0 are x = (-1 + √13)/6 or x = (-1 - √13)/6, that is x ≈ -0.76759188 and x ≈ 0.43425855.

x² +x += 0
−3x2−x+1=0⇒x=−1+136, −1−136-3x^2 - x + 1 = 0 \Rightarrow x = \frac{-1 + \sqrt{13}}{6},\ \frac{-1 - \sqrt{13}}{6}
Discriminant D = 13 · x ≈ -0.767592, 0.434259

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

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Graph of y = -3x² - x + 1-3-2-112-25-20-15-10-5x = -0.76759188x = 0.43425855vertexy = -3x² - x + 1
Graph of y = -3x² - x + 1. The curve crosses the x-axis at x = -0.76759188 and x = 0.43425855; the vertex is at (-0.16666667, 1.0833333).

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 = -1, c = 1.

    −3x2−x+1=0-3x^2 - x + 1 = 0
  2. 2

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

    D=(−1)2−4⋅(−3)⋅1=1+12=13D = (-1)^2 - 4\cdot(-3)\cdot1 = 1 + 12 = 13
  3. 3

    D is positive, so there are two different real solutions.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=1±13−6x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{13}}{-6}
  5. 5

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

    x=−1+136orx=−1−136x = \frac{-1 + \sqrt{13}}{6}\quad\text{or}\quad x = \frac{-1 - \sqrt{13}}{6}
  6. 6

    As decimals: x ≈ -0.76759188 and x ≈ 0.43425855.

  7. 7

    Check numerically: for x ≈ -0.76759188 the left side is 0 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -3x^2 - x + 1 = 0\ \Rightarrow\ x = \frac{-1 + \sqrt{13}}{6},\ \frac{-1 - \sqrt{13}}{6} \]
\[ -3x^2 - x + 1 = 0 \]
\[ D = (-1)^2 - 4\cdot(-3)\cdot1 = 1 + 12 = 13 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{1 \pm \sqrt{13}}{-6} \]
\[ x = \frac{-1 + \sqrt{13}}{6}\quad\text{or}\quad x = \frac{-1 - \sqrt{13}}{6} \]

The parabola y = -3x² - x + 1

xy
-3.1667-25.917267
-2.1667-10.917067
-1.1667-1.9168667
-0.16671.0833333
0.8333-1.9164667
1.8333-10.916267
2.8333-25.916067

Questions

What are the solutions of -3x² - x + 1 = 0?

The solutions of -3x² - x + 1 = 0 are x = (-1 + √13)/6 or x = (-1 - √13)/6, that is x ≈ -0.76759188 and x ≈ 0.43425855.

Does -3x² - x + 1 = 0 factor?

Not over the integers, because the discriminant 13 is not a perfect square. Use the quadratic formula.

What is the vertex of y = -3x² - x + 1?

(-1/6, 13/12). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = -1/3, product = c/a = -1/3.

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