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Solve x² + 6x + 3 = 0

The solutions of x² + 6x + 3 = 0 are x = -3 + √6 or x = -3 - √6, that is x ≈ -0.55051026 and x ≈ -5.4494897.

x² +x += 0
x2+6x+3=0⇒x=−3+6, −3−6x^2 + 6x + 3 = 0 \Rightarrow x = -3 + \sqrt{6},\ -3 - \sqrt{6}
Discriminant D = 24 · x ≈ -0.55051, -5.44949

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

Graph of y = x² + 6x + 3-7-6-5-4-3-2-11-55101520x = -0.55051026x = -5.4494897vertexy = x² + 6x + 3
Graph of y = x² + 6x + 3. The curve crosses the x-axis at x = -0.55051026 and x = -5.4494897; the vertex is at (-3, -6).

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 = 6, c = 3.

    x2+6x+3=0x^2 + 6x + 3 = 0
  2. 2

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

    D=62−4⋅1⋅3=36−12=24D = 6^2 - 4\cdot1\cdot3 = 36 - 12 = 24
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−6±242x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{24}}{2}
  5. 5

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

    x=−3+6orx=−3−6x = -3 + \sqrt{6}\quad\text{or}\quad x = -3 - \sqrt{6}
  6. 6

    As decimals: x ≈ -0.55051026 and x ≈ -5.4494897.

  7. 7

    Check numerically: for x ≈ -0.55051026 the left side is -8.8817842e-16 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 + 6x + 3 = 0\ \Rightarrow\ x = -3 + \sqrt{6},\ -3 - \sqrt{6} \]
\[ x^2 + 6x + 3 = 0 \]
\[ D = 6^2 - 4\cdot1\cdot3 = 36 - 12 = 24 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{24}}{2} \]
\[ x = -3 + \sqrt{6}\quad\text{or}\quad x = -3 - \sqrt{6} \]

The parabola y = x² + 6x + 3

xy
-63
-5-2
-4-5
-3-6
-2-5
-1-2
03

Questions

What are the solutions of x² + 6x + 3 = 0?

The solutions of x² + 6x + 3 = 0 are x = -3 + √6 or x = -3 - √6, that is x ≈ -0.55051026 and x ≈ -5.4494897.

Does x² + 6x + 3 = 0 factor?

Not over the integers, because the discriminant 24 is not a perfect square. Use the quadratic formula.

What is the vertex of y = x² + 6x + 3?

(-3, -6). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -6, product = c/a = 3.

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    Solve x² + 6x + 3 = 0: x = -3 + √6 or x = -3 - √6 (steps) | WordGPT