Triangle abc is such that ab=4 and ac=8 if m is the midpoint of bc and am=3, what is the length of bc?

This is an interesting question. I chose to tackle it using the Law of Cosines.
  AC² = AB² + BC² - 2·AB·BC·cos(B)
  AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
  AC² - 2·AM² = -AB² + BC² - 2·MB²

We know that MB = BC/2. When we substitute the given information, we have
  8² - 2·3² = -4² + BC² - BC²/2
  124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
  2√31 = BC ≈ 11.1355

Something that is true about that is that it has numbers

Use a graph site then you will find your answer