Question 973230
Find three numbers in continued proportion such that their sum is 14 and sum 
of their squares is 84.
<pre>
{{{system("x:y" = "y:z", x+y+z = 14, x^2+y^2+z^2 = 84)}}}

{{{system(x/y = y/z, x+y+z = 14, x^2+y^2+z^2 = 84)}}}

{{{system(xz = y^2, x+y+z = 14, x^2+y^2+z^2 = 84)}}}
                           __
Substitute xz for y² and ±&#8730;xz for y in 3rd and 2nd equations:

{{{system(x +- sqrt(xz)+ z = 14, x^2+xz+z^2 = 84)}}}
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Solve 1st equation for ±&#8730;xy

{{{system("" +- sqrt(xz)= 14-x-z, x^2+xz+z^2 = 84)}}}

Square both sides of 1st equation

{{{system(xz= 196+x^2+z^2-28x-28z+2xz, x^2+xz+z^2 = 84)}}}

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

{{{system(-x^2-xz-z^2=196-28x-28z, x^2+xz+z^2 = 84)}}}

Add the two equations

{{{0=280-28x-28z}}}

Divide through by 28

{{{0=10-x-z}}}

{{{x+z=10}}}

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

{{{system(x+y+z=14,x+z=10)}}}

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</pre>