How do you find the square root of 81?

3 Answers
Jul 6, 2016

Answer:

The square root of 81 is #+-9#

Explanation:

The square root of 81 is #sqrt81#

#+-sqrt(9*9)#

#+-9#

Jul 6, 2016

Answer:

It is #\pm9#.

Explanation:

#81=9*9# then the square root of #sqrt(81)=9#.
Because the double multiplication for the same sign is always positive, the square root is also valid with the other sign #81=(-9)*(-9)# then #sqrt(81)=-9# and we can say that #sqrt(81)=\pm9#.

What if we do not know the value? There is an algorithm to calculate the square root quite simple (the Babylonian algorithm).

We want to calculate the square root of #81#, we first guess a possible value. For example we know that #10*10=100# and #7*7=49#, then we can imagine that the square root of #81# is between #7# and #10#, we can imagine for example #8.3#.

Then the algorithm tell us that the next step of approximation is given by

#x_(n+1)=1/2(x_n+alpha/x_n)# where alpha is the number that we are searching the square root (for us #alpha=81#) and #x_n# is our guess. Our first guess is #x_1=8.3#, then the next step of approximation is

#x_2=1/2(8.3+81/8.3)=9.029518#.

I continue with the next approximation

#x_3=1/2(9.029518+81/9.029518)=9.0000482#

#x_4=1/2(9.0000482+81/9.0000482)=9.00000000013#

#x_5=1/2(9.00000000013+81/9.00000000013)=9# (my computer does not approximate more).

As you can see in few steps the algorithm converges to the square root. Of course, because the previous discussion about the positive and negative sign, also the same result with the minus sign is a solution of the square root.

For me it is very fascinating the fact that I can write on Socratic.org, on Internet, an answer to someone that I do not know and that probably lives in a place that I will never see in my life, an algorithm to calculate the square root of a number invented 4000 years ago by some very clever Babylonian. Mathematics is the most universal language, trough space and time.

May 30, 2017

Answer:

#sqrt81 =9#

Explanation:

The square root of a number is that number which, when multiplied by itself will give the original number:

#sqrty = sqrt(? xx ?) = ?" "sqrt25= sqrt (5xx5) = 5#

#sqrt(x^2)= sqrt(x xx x) = x" "sqrt (x^6y^8) = x^3y^4#

To find any root of a number, write the radicand as the product of its prime factors:

#sqrt81 = sqrt(3xx3xx3xx3)#

You can now understand this in different ways:

  • #sqrt81 = sqrt((3xx3)xx(3xx3)) = 3xx3 = 9#

  • #sqrt81 = sqrt((3xx3)xx(3xx3)) = sqrt(9xx9) = 9#

  • #sqrt81 = sqrt(3^4) = 3^(4div2) = 3^2 =9#

Note: #sqrt"radicand"#