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Solve x² + 12x + 1 = 0

The solutions of x² + 12x + 1 = 0 are x = -6 + √35 or x = -6 - √35, that is x ≈ -0.083920217 and x ≈ -11.91608.

x² +x += 0
x2+12x+1=0⇒x=−6+35, −6−35x^2 + 12x + 1 = 0 \Rightarrow x = -6 + \sqrt{35},\ -6 - \sqrt{35}
Discriminant D = 140 · x ≈ -0.0839202, -11.9161

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

Graph of y = x² + 12x + 1-14-12-10-8-6-4-22-40-30-20-1010203040x = -0.083920217x = -11.91608vertexy = x² + 12x + 1
Graph of y = x² + 12x + 1. The curve crosses the x-axis at x = -0.083920217 and x = -11.91608; the vertex is at (-6, -35).

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 = 12, c = 1.

    x2+12x+1=0x^2 + 12x + 1 = 0
  2. 2

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

    D=122−4⋅1⋅1=144−4=140D = 12^2 - 4\cdot1\cdot1 = 144 - 4 = 140
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−12±1402x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{140}}{2}
  5. 5

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

    x=−6+35orx=−6−35x = -6 + \sqrt{35}\quad\text{or}\quad x = -6 - \sqrt{35}
  6. 6

    As decimals: x ≈ -0.083920217 and x ≈ -11.91608.

  7. 7

    Check numerically: for x ≈ -0.083920217 the left side is 9.9920072e-16 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 + 12x + 1 = 0\ \Rightarrow\ x = -6 + \sqrt{35},\ -6 - \sqrt{35} \]
\[ x^2 + 12x + 1 = 0 \]
\[ D = 12^2 - 4\cdot1\cdot1 = 144 - 4 = 140 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{140}}{2} \]
\[ x = -6 + \sqrt{35}\quad\text{or}\quad x = -6 - \sqrt{35} \]

The parabola y = x² + 12x + 1

xy
-9-26
-8-31
-7-34
-6-35
-5-34
-4-31
-3-26

Questions

What are the solutions of x² + 12x + 1 = 0?

The solutions of x² + 12x + 1 = 0 are x = -6 + √35 or x = -6 - √35, that is x ≈ -0.083920217 and x ≈ -11.91608.

Does x² + 12x + 1 = 0 factor?

Not over the integers, because the discriminant 140 is not a perfect square. Use the quadratic formula.

What is the vertex of y = x² + 12x + 1?

(-6, -35). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -12, product = c/a = 1.

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    Solve x² + 12x + 1 = 0: x = -6 + √35 or x = -6 - √35 (steps) | WordGPT