SOLUTION: Find three numbers in continued proportion such that their sum is 14 and sum of their squares is 84.

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Question 973230: Find three numbers in continued proportion such that their sum is 14 and sum
of their squares is 84.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Find three numbers in continued proportion such that their sum is 14 and sum
of their squares is 84.
system%28%22x%3Ay%22+=+%22y%3Az%22%2C+x%2By%2Bz+=+14%2C+x%5E2%2By%5E2%2Bz%5E2+=+84%29

system%28x%2Fy+=+y%2Fz%2C+x%2By%2Bz+=+14%2C+x%5E2%2By%5E2%2Bz%5E2+=+84%29

system%28xz+=+y%5E2%2C+x%2By%2Bz+=+14%2C+x%5E2%2By%5E2%2Bz%5E2+=+84%29
                           __
Substitute xz for y² and ±√xz for y in 3rd and 2nd equations:

system%28x+%2B-+sqrt%28xz%29%2B+z+=+14%2C+x%5E2%2Bxz%2Bz%5E2+=+84%29
                         __
Solve 1st equation for ±√xy

system%28%22%22+%2B-+sqrt%28xz%29=+14-x-z%2C+x%5E2%2Bxz%2Bz%5E2+=+84%29

Square both sides of 1st equation

system%28xz=+196%2Bx%5E2%2Bz%5E2-28x-28z%2B2xz%2C+x%5E2%2Bxz%2Bz%5E2+=+84%29

Get -x²-xz-z² on the left of the 1st equation:

system%28-x%5E2-xz-z%5E2=196-28x-28z%2C+x%5E2%2Bxz%2Bz%5E2+=+84%29

Add the two equations

0=280-28x-28z

Divide through by 28

0=10-x-z

x%2Bz=10

Put that with original equation x+y+z = 14

system%28x%2By%2Bz=14%2Cx%2Bz=10%29

Subtract the second equation from the first:

y=4, and since x+z=10, z=10-x

Substitute those in the original equation x²+y²+z²=84 

        x²+y²+z² = 84
   x²+4²+(10-x)² = 84
x²+16+100-20x+x² = 84
     2x²-20x+116 = 84
      2x²-20x+32 = 0

Divide through by 2

       x²-10x+16 = 0
      (x-8)(x-2) = 0
   x-8 = 0;    x-2 = 0
     x = 8;      x = 2
     z = 10-x;   z = 10-x
     z = 10-8;   z = 10-2
     z = 2;      z = 8

So there seems to be two solutions:

(x,y,z) = (8,4,2)  
(x,y,z) = (2,4,8)

Checking: 8:4 = 4:2  that checks
          2:4 = 4:8  that checks

Actually there is really only 1 solution because

if x:y = y:z, then z:y = y:x and vice-versa.

Edwin