Question

Estimate each square root to the nearest hundredth of 32. Show your solution!​

Answer 1

Answer:

${ \rm{ \sqrt{32} = \sqrt{2 \times 16} }} \\ \\ = { \rm{ \sqrt{2 \times 4 \times 4} }} \\ \\ = { \rm{ \sqrt{2} \times( \sqrt{4}) {}^{2} }} \\ \\ = { \rm{ \sqrt{2} \times 4 }} \\ \\ = { \rm{4 \sqrt{2} }} \\ \\ { \rm{ \sqrt{32} = 5.656854....}} \\ \\ { \boxed{ \rm{ \sqrt{32} = 5.66 \: }}}$

Answer 2

9514 1404 393

Answer:

  5.66

Step-by-step explanation:

For an integer x = s² +r, where s and r are also integers, a first approximation of the square root of x will be ...

  y = √x ≈ s +r/(2s+1)

An iterative improvement on the estimate can be had by ...

  y' = s + r/(s +y)

Here, this means a first estimate can be obtained as ...

  32 = 5² +7   ⇒   s = 5, r = 7

  √32 ≈ y1 = s +r/(2s +1) = 5 7/11

Then an iterative improvement will be ...

  y2 = s +r/(s +y1) = 5 +7/(10 7/11) = 5 77/117 ≈ 5.658 ≈ 5.66

_____

Additional comment

The above iterator will add about 1 decimal place for each iteration. The Babylonian method will roughly double the number of good decimal places with each iteration. For an approximate root of p/q, the next iteration using that method would be ...

  p'/q' = (x·q² +p²)/(2pq)

Then our above estimate of 5 7/11 for the root of 32 would be refined to ...

  p'/q' = (32·11² +62²)/(2·11·62) = 7716/1364 = 5 224/341 ≈ 5.65689.... This compares favorably to the first digits of the actual root: 5.65685....