How do you find the square root of 81?

3 Answers
Jul 6, 2016

The square root of 81 is +-9

Explanation:

The square root of 81 is sqrt81

+-sqrt(9*9)

+-9

Jul 6, 2016

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

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"