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Solve 4x² - 7x + 11 = 0

4x² - 7x + 11 = 0 has no real solutions. The two complex solutions are x = (7 + √127 i)/8 and x = (7 - √127 i)/8.

x² +x += 0
4x2−7x+11=0⇒x=7+127 i8, 7−127 i8 (complex)4x^2 - 7x + 11 = 0 \Rightarrow x = \frac{7 + \sqrt{127}\, i}{8},\ \frac{7 - \sqrt{127}\, i}{8}\ \text{(complex)}
Discriminant D = -127

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

Graph of y = 4x² - 7x + 11-3-2-1123410203040506070vertexy = 4x² - 7x + 11
Graph of y = 4x² - 7x + 11. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.875, 7.9375).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 4, b = -7, c = 11.

    4x2−7x+11=04x^2 - 7x + 11 = 0
  2. 2

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

    D=(−7)2−4⋅4⋅11=49−176=−127D = (-7)^2 - 4\cdot4\cdot11 = 49 - 176 = -127
  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=7±−1278x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{-127}}{8}
  5. 5

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

    x=7+127 i8orx=7−127 i8x = \frac{7 + \sqrt{127}\, i}{8}\quad\text{or}\quad x = \frac{7 - \sqrt{127}\, i}{8}
LaTeX source of the formulas on this page
\[ 4x^2 - 7x + 11 = 0\ \Rightarrow\ x = \frac{7 + \sqrt{127}\, i}{8},\ \frac{7 - \sqrt{127}\, i}{8} \]
\[ 4x^2 - 7x + 11 = 0 \]
\[ D = (-7)^2 - 4\cdot4\cdot11 = 49 - 176 = -127 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{-127}}{8} \]
\[ x = \frac{7 + \sqrt{127}\, i}{8}\quad\text{or}\quad x = \frac{7 - \sqrt{127}\, i}{8} \]

The parabola y = 4x² - 7x + 11

xy
-2.12543.9375
-1.12523.9375
-0.12511.9375
0.8757.9375
1.87511.9375
2.87523.9375
3.87543.9375

Questions

What are the solutions of 4x² - 7x + 11 = 0?

4x² - 7x + 11 = 0 has no real solutions. The two complex solutions are x = (7 + √127 i)/8 and x = (7 - √127 i)/8.

Does 4x² - 7x + 11 = 0 factor?

Not over the integers, because the discriminant -127 is negative. Use the quadratic formula.

What is the vertex of y = 4x² - 7x + 11?

(7/8, 127/16). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 7/4, product = c/a = 11/4.

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