How do you find a square root of a decimal?

Question

Genesis Allen · Accepted Answer

undefined To find the square root of a decimal number, you can use a calculator or a mathematical method such as long division or the Newton-Raphson method. Simply input the decimal number into the calculator or follow the steps of the chosen method to calculate the square root.

Abigail Allen · Answer

We can use special long division method (other than using a calculator) described below.

Let us consider a number #0.04783#, which is not a perfect square, to find the square root of #0.04783#, we should do a special long division, where we pair, the numbers in two, starting from decimal point in either direction. Here, however, we have chosen a number which has #0# to the left of decimal point. To the right of decimal the pairs formed are #0. color(red)(04)color(blue)(78)color(red)(30)#. We write the number in long form of division as shown below. #color(white)(xx)ul(0. color(white)(x)2color(white)(xx)1color(white)(xx)8color(white)(xx)7color(white)(xx)0)# #ul2|0. color(red)(04)color(white)(x)color(blue)(78)color(white)(x)color(red)(30)color(white)(x)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)# #color(white)(xxxx)ul(04)color(white)(X)darr# #color(red)(4)1|color(white)(X)00color(white)(x)78# #color(white)(xxxx)ul(00color(white)(x)41)# #color(white)(x)color(red)(42)8|color(white)(xx)37color(white)(.)30# #color(white)(xxxxxx.)ul(34color(white)(.)24)# #color(white)(xx)color(red)(436)7|color(white)(.)3color(white)(x)06color(white)(.)00# #color(white)(xxxxxXx)ul(3color(white)(.)05color(white)(.)69)# #color(white)(xx)color(red)(4374)0|color(white)(x)color(white)(.)00color(white)(.)31color(white)(.)00# Here we have first pair is #04# and the number whose square is just equal or less than it is #02#, so we get #2# and write its square #04# below #04#. The difference is #00# and now we bring down next two digits #78#. As a divisor we first write double of #2# i.e. #4# and then find a number #x# so that #4x# (here #x# stands for single digit in units place) multiplied by #x# is just less than the number, here #0078#. We find for #x=1#, we have #41xx1=41# and get the difference as #37#. Now as we have a remainder of #37#, we bring #30# making the number #3730#. Observe that we had brought as divisor #2xx2=4#, but this time we have #21#, whose double is #42#, and so we make the divisor as #42x# and identify an #x# so that #42x# multiplied by #x# is just less than #3730#. This number is just #8#, as making it #9# will make the product #429xx9=3861>3760#. With #8#, we get #428xx8=3424# and remainder is #306#. Next, we do not have anything, but for accuracy we can bring #00#. The divisor will now be #218xx2=436# and identify an #x# so that #436x# multiplied by #x# is just less than #30600#. The number we get is #7#. We continue to do similarly and find remainder is just #31#, while we will have #2187xx2=4374# i.e. #4374x# and this #x# ought to be zero. It is evident that we have reached desired accuracy and #sqrt0.04783~~0.2187# Another example is here.