SOLUTION: Can the following expression ((X+5)^0.5) be multiplied out like the following expression (x+5)^2. Thank you for the consideration.

Algebra ->  Square-cubic-other-roots -> SOLUTION: Can the following expression ((X+5)^0.5) be multiplied out like the following expression (x+5)^2. Thank you for the consideration.       Log On


   

Question 1189036: Can the following expression ((X+5)^0.5) be multiplied out like the following expression (x+5)^2. Thank you for the consideration.

Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

%28%28x%2B5%29%5E0.5%29+%2A+%28x%2B5%29%5E2+
%28%28x%2B5%29%5E%281%2F2%29%29+%2A+%28x%2B5%29%5E2+....apply The Product Rule for Exponents: a%5Em%2Aa%5En+=+a%5E%28m+%2B+n%29
%28x%2B5%29%5E%281%2F2%2B2%29+
%28x%2B5%29%5E%285%2F2%29+

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


I very much think the other tutor completely missed the point of your question....

Yes, you can "multiply out" the expression ((x+5)^(0.5)) using binomial expansion.

Many students will be familiar with binomial expansion when the power is an integer; few students will be familiar with it when the power is not an integer.

Because the exponent is not an integer, the expansion does not terminate; it has an infinite number of terms. The first few terms will give good approximations.

The key to understanding binomial expansion with non-integer powers is to look at how binomial coefficients are calculated.

For all n, including non-integer values of n...

C%28n%2C0%29+=+1

For all other binomial coefficients C(n,r), the coefficient can be calculated as a fraction with "r" numbers in both numerator and denominator, starting with n and decrementing by 1 in the numerator, and starting with r and decrementing by 1 in the denominator. That gives us...

C%28n%2C1%29+=+n%2F1+=+n
C%28n%2C2%29+=+%28%28n%29%28n-1%29%29%2F%282%2A1%29
C%28n%2C3%29+=+%28%28n%29%28n-1%29%28n-2%29%29%2F%283%2A2%2A1%29
etc....

For example for n=3, we get what (I hope) are the familiar coefficients for n=3:

C%283%2C0%29+=+1
C%283%2C1%29+=+%283%2F1%29+=+3
C%283%2C2%29+=+%28%283%2A2%29%2F%282%2A1%29%29+=+6%2F2+=+3
C%283%2C3%29+=+%28%283%2A2%2A1%29%2F%283%2A2%2A1%29%29+=+6%2F6+=+1

And using the same method for calculating the coefficients for n=0.5, we get the unfamiliar coefficients for a binomial expansion to the 0.5 power:

C%280.5%2C0%29+=+1
C%280.5%2C1%29+=+%280.5%2F1%29+=+1%2F2
C%280.5%2C2%29+=+%28%28%280.5%29%28-0.5%29%29%2F%282%2A1%29%29+=+-1%2F8
C%280.5%2C3%29+=+%28%28%280.5%29%28-0.5%29%28-1.5%29%29%2F%283%2A2%2A1%29%29+=+1%2F16

Now we can write out the first few terms of the binomial expansion of (x+5)^(-0.5). We have the binomial coefficients; and from one term to the next the power of x decreases by 1 and the power of 5 increases by 1:



Evaluating that expression for values of x that should produce whole number results, we find

   x  x+5  (x+5)^(0.5)  f(x)  (4 decimal places)
  ------------------------------
   4   9        3       3.1035
  11  16        4       4.0042
  20  25        5       5.0006
  31  36        6       6.0001
  44  49        7       7.0000
  59  64        8       8.0000

For small values of x, where this approximation is not very accurate, using 2 or 3 more terms of the binomial expansion would improve the accuracy.