 In his commentary on the nine chapters, Liu Hui, 3rd century AD, encounters a problem we would write as x squared plus 34x equals 71,000. He tells the reader to extract the root, but gives no details. What did he mean? To understand what Liu Hui meant, let's compare two problems. Find the square root of 85,849, and solve x squared equals 85,849. And the thing to recognize is that they're the same problem. And that means the root procedure is also a procedure for solving quadratic equations. So let's solve x squared plus 35x equals 1,334. So we'll set up our counting board. The number 1,334 is going to be our top row. What we can call the sides, that's the coefficient of x, 35 is our second row. For the squares, that's the coefficient of x squared, 1 is going to be our bottom row. Now since this is a quadratic and analogous to the square root, we'll shift the squares two places at a time. Now if you go back to the root procedure, in the root procedure, whatever the placeholder shifts two places, the ding fa shifts one. And so we shift the sides one place. And now we take our guess. The product of the guess and the placeholder will be added to the 3 directly above the placeholder. Whatever that sum is, times the guess must be less than 13. And so we'll guess 2 and perform the first pass. So guess times squares and add to sides, guess times sides, and subtract. Now in the square root procedure we would have doubled the partial root and replaced the sides. But because we already had some sides there, we have to do something a little bit differently. We have to do what we can call a second pass. We're going to take guess times squares and add to the sides. And now we're going to shift everything over. The sides move one place and the squares move two places. And now we go back to the beginning. And the guess times the squares will be added to the sides, and that means the guess times the sides must be less than this remainder 234. And so we guess about 3. And so we take the first pass, guess times squares and add to sides, guess times sides, and subtract. And since we've exhausted the root, we've found our solution, 23. As with our square root procedure, we can also find decimal approximations to any desired degree of accuracy. For example, let's have it to solve x squared plus 10x equals 50, and let's express our answer to 2 decimal places. So we'll set down our number, 50, our sides, coefficient of x, 10, and our squares, coefficient of x squared, 1. If we don't need to move the placeholder, our root will have 1 whole number digit. So we'll guess 4 and make our first pass. Guess times squares and add to sides, then guess times sides and subtract, but we can't, we have too much. And that means we need to reset and guess lower. So this time, we'll guess 3. So guess times squares and add to sides, guess times sides, and subtract. We'll make our second pass, guess times squares and add to sides, and shift. And so our next guess is, we'll have about 6. So guess times squares and add to sides, guess times sides, and subtract. And now for our second pass, guess times squares and add to sides, and shift. And once more until the breach, we guess 6. So guess times squares and add to sides, guess times sides, and subtract. And so far the digits of the root are 3, 6, 6. And since we determined the root has 1 whole number digit, this allows us to place the decimal. And so we have our answer to 2 decimal places, 3.66.