Question 1098559
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x^2+xy+yz+zx=30 
y^2+xy+yz+zx=15
z^2+xy+yz+zx=18 

find x^2+y^2+z^2
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<U>Solution</U>


<pre>
<U>Step 1</U>.  Factor left side of each equation. You will get

    x^2+xy+yz+zx = 30    is the same as    (x+y)*(x+z) = 30    (1)
    y^2+xy+yz+zx = 15    is the same as    (x+y)*(y+z) = 15    (2)
    z^2+xy+yz+zx = 18    is the same as    (x+z)*(y+z) = 18    (3)



<U>Step 2</U>.  Multiply equations (1), (2) and (3)  (left sides and right sides). You will get


    {{{(x+y)^2*(x+z)^2*(y+z)^2}}} = 30*15*18 = (15*2)*15*(2*3^2) = {{{2^2*3^2*15^2}}} = {{{(2*3*15)^2}}} = {{{90^2}}}.

    Take the square root of both sides.  You will get  (x+y)*(x+z)*(y+z) = +/- {{{sqrt(90^2)}}},   or

    (x+y)*(x+z)*(y+z) = +/- 90.    (4)



<U>Step 3</U>.  First, let us consider the case

    (x+y)*(x+z)*(y+z) = 90.        (5)


    Divide eq(5) by eq(1) (both sides).  You will get    y + z = 3.       (6)    (3 = 90/30)

    Divide eq(5) by eq(2) (both sides).  You will get    x + z = 6.       (7)    (6 = 90/15)

    Divide eq(5) by eq(3) (both sides).  You will get    x + y = 5.       (8)    (5 = 90/18)



<U>Step 4</U>.  To solve the system (6), (7), (8), add all three equation (6), (7) and (8) (both sides). You will get

    
    2x + 2y + 2z = 3 + 6 + 5 = 14.  It implies           x + y + z = 7.   (9)


    Now   subtract eq(6) from eq(9)  (both sides).   You will get  x = 7-3 = 4.

    Next, subtract eq(7) from eq(9)  (both sides).   You will get  y = 7-6 = 1.

    Next, subtract eq(7) from eq(9)  (both sides).   You will get  z = 7-5 = 2.


    So, (x,y,z) = (4,1,2) is the solution in this case.


<U>Step 5</U>.  Now let us consider the case

    (x+y)*(x+z)*(y+z) = -90.        


    Doing by the same way, you will obtain the solution  (x,y,z) = (-4,-1,-2)  in this case.


<U>Step 6</U>.  <U>Check</U>.  It is enough to check the positive solution only.

     
    4^2 + 4*1 + 4*2 + 1*2 = 30.     ! Correct !
    1^2 + 4*1 + 4*2 + 1*2 = 15.     ! Correct !
    2^2 + 4*1 + 4*2 + 1*2 = 18.     ! Correct !


<U>Answer</U>.  The given system has two and only two solutions  (x,y,z) = (4,1,2)  and  (x,y,z) = (-4,-1,-2).

         Therefore,  {{{x^2 + y^2 + z^2}}} = {{{4^2 +1^2 + 2^2}}} = 21.
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