Welcome to a lesson on approximating square roots. Using division this process
will allow you to approximate a square root without the use of a calculator.
Let's begin by approximating the square root of 23. Since we know the square root
of 16 equals 4 since 4 squared equals 16, and the square root of 25 equals 5 since
5 squared equals 25, It is reasonable to guess the square root of 23 is approximately
5, since 23 is closer to 25 than it is to 16. Let's let the first approximation of
the square root of 23 be equal to 5. The next step is to divide 23, the radicand, by the
first approximation of 5. Notice 23 divided by 5 equals 4.6. The next step is
to make a new approximation between the quotient of 4.6 and the original
approximation of 5, and then we divide this into 23 again. Let's guess 4.8
because 4.8 is between 4.6 and 5. Notice 23 divided by 4.8 is approximately four
point seven nine and four point seven nine is very close to our approximation
of four point eight, so we could repeat this process by now making a new
approximation between four point seven nine and four point eight but notice how
the square root of twenty-three is approximately four point seven nine
since the square of four point seven nine is equal to 22 point nine four four
one, which is very close to 23. So as we go through this process as the quotient
becomes closer and closer to our approximation, we have more and more
accuracy. Let's look at a second example. Let's now approximate the square root of
68. Since we know the square root of 64 equals eight and the square root of 81
equals nine it is reasonable to guess the square root of 68 is approximately
eight since 68 is closer to sixty-four than it is to 81. Let's let the first
approximation for the square root of 68 be eight. Step two, we divide the radicand
of 68 by eight. Sixty eight divided by 8 equals eight
point five. Step three, we will now make a new approximation between the quotient
of eight point five and the original guess of eight, and then we divide this
into 68. Let's guess eight point two five because eight point two five is between
eight point five and eight and now we divide 68 by eight point two five, which
gives us approximately eight point two four. Again notice how eight point two
four is very close to our approximation of eight point two five and therefore
eight point two four is a good approximation for the square root of 68.
Again we could repeat this process that making a new approximation between eight
point two four and eight point two five but the square root of 68 is approximately
eight point two four since the square of eight point two four equals sixty seven
point eight nine, which again is very close to 68. Now let's take a look at why
this process works. In example one we were approximating the square root of
twenty-three so if you let n equal the exact value of
the square root of twenty three, and then we square both sides of the equation we
get the equation n squared equals twenty three where n is greater than or equal
to zero. From here if we divide both sides of the equation by n and simplify,
we end up with n equals 23 divided by n. So if we knew the exact value of the
square root of twenty three, the exact value would also be equal to this
quotient, which means the real value of the square root of 23 must be somewhere
between an approximate value of the square root of twenty-three and twenty
three divided by the approximate value. Similarly for the square root of 68 the
real value of the square root of sixty eight must be somewhere between an
approximate value of the square root of 68 and 68 divided by the approximate
value. This is the reason why this division algorithm works to approximate
square roots. I hope you found this helpful.