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Solve 2x² + 11x + 1 = 0

The solutions of 2x² + 11x + 1 = 0 are x = (-11 + √113)/4 or x = (-11 - √113)/4, that is x ≈ -0.092463547 and x ≈ -5.4075365.

x² +x += 0
2x2+11x+1=0⇒x=−11+1134, −11−11342x^2 + 11x + 1 = 0 \Rightarrow x = \frac{-11 + \sqrt{113}}{4},\ \frac{-11 - \sqrt{113}}{4}
Discriminant D = 113 · x ≈ -0.0924635, -5.40754

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

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Graph of y = 2x² + 11x + 1-7-6-5-4-3-2-112-20-1010203040x = -0.092463547x = -5.4075365vertexy = 2x² + 11x + 1
Graph of y = 2x² + 11x + 1. The curve crosses the x-axis at x = -0.092463547 and x = -5.4075365; the vertex is at (-2.75, -14.125).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 2, b = 11, c = 1.

    2x2+11x+1=02x^2 + 11x + 1 = 0
  2. 2

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

    D=112−4⋅2⋅1=121−8=113D = 11^2 - 4\cdot2\cdot1 = 121 - 8 = 113
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−11±1134x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{113}}{4}
  5. 5

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

    x=−11+1134orx=−11−1134x = \frac{-11 + \sqrt{113}}{4}\quad\text{or}\quad x = \frac{-11 - \sqrt{113}}{4}
  6. 6

    As decimals: x ≈ -0.092463547 and x ≈ -5.4075365.

  7. 7

    Check numerically: for x ≈ -0.092463547 the left side is 1.5543122e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ 2x^2 + 11x + 1 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{113}}{4},\ \frac{-11 - \sqrt{113}}{4} \]
\[ 2x^2 + 11x + 1 = 0 \]
\[ D = 11^2 - 4\cdot2\cdot1 = 121 - 8 = 113 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{113}}{4} \]
\[ x = \frac{-11 + \sqrt{113}}{4}\quad\text{or}\quad x = \frac{-11 - \sqrt{113}}{4} \]

The parabola y = 2x² + 11x + 1

xy
-5.753.875
-4.75-6.125
-3.75-12.125
-2.75-14.125
-1.75-12.125
-0.75-6.125
0.253.875

Questions

What are the solutions of 2x² + 11x + 1 = 0?

The solutions of 2x² + 11x + 1 = 0 are x = (-11 + √113)/4 or x = (-11 - √113)/4, that is x ≈ -0.092463547 and x ≈ -5.4075365.

Does 2x² + 11x + 1 = 0 factor?

Not over the integers, because the discriminant 113 is not a perfect square. Use the quadratic formula.

What is the vertex of y = 2x² + 11x + 1?

(-11/4, -113/8). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -11/2, product = c/a = 1/2.

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    Solve 2x² + 11x + 1 = 0: x = (-11 + √113)/4 or x = (-11 - √113)/4 (steps) | WordGPT