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Solve 5x² + 11x + 8 = 0

5x² + 11x + 8 = 0 has no real solutions. The two complex solutions are x = (-11 + √39 i)/10 and x = (-11 - √39 i)/10.

x² +x += 0
5x2+11x+8=0⇒x=−11+39 i10, −11−39 i10 (complex)5x^2 + 11x + 8 = 0 \Rightarrow x = \frac{-11 + \sqrt{39}\, i}{10},\ \frac{-11 - \sqrt{39}\, i}{10}\ \text{(complex)}
Discriminant D = -39

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

Graph of y = 5x² + 11x + 8-5-4-3-2-11220406080vertexy = 5x² + 11x + 8
Graph of y = 5x² + 11x + 8. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (-1.1, 1.95).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 5, b = 11, c = 8.

    5x2+11x+8=05x^2 + 11x + 8 = 0
  2. 2

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

    D=112−4⋅5⋅8=121−160=−39D = 11^2 - 4\cdot5\cdot8 = 121 - 160 = -39
  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±−3910x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{-39}}{10}
  5. 5

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

    x=−11+39 i10orx=−11−39 i10x = \frac{-11 + \sqrt{39}\, i}{10}\quad\text{or}\quad x = \frac{-11 - \sqrt{39}\, i}{10}
LaTeX source of the formulas on this page
\[ 5x^2 + 11x + 8 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{39}\, i}{10},\ \frac{-11 - \sqrt{39}\, i}{10} \]
\[ 5x^2 + 11x + 8 = 0 \]
\[ D = 11^2 - 4\cdot5\cdot8 = 121 - 160 = -39 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{-39}}{10} \]
\[ x = \frac{-11 + \sqrt{39}\, i}{10}\quad\text{or}\quad x = \frac{-11 - \sqrt{39}\, i}{10} \]

The parabola y = 5x² + 11x + 8

xy
-4.146.95
-3.121.95
-2.16.95
-1.11.95
-0.16.95
0.921.95
1.946.95

Questions

What are the solutions of 5x² + 11x + 8 = 0?

5x² + 11x + 8 = 0 has no real solutions. The two complex solutions are x = (-11 + √39 i)/10 and x = (-11 - √39 i)/10.

Does 5x² + 11x + 8 = 0 factor?

Not over the integers, because the discriminant -39 is negative. Use the quadratic formula.

What is the vertex of y = 5x² + 11x + 8?

(-11/10, 39/20). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -11/5, product = c/a = 8/5.

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