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  Three real numbers x, y, z are such that x2 + 6y = -􀀀17, y2 + 4z = 1 and z2 + 2x = 2. What
is the value of x2 + y2 + z2?
solve this if you can
x2 =x^2 same for others

 Sep 26, 2015

Best Answer 

 #1
avatar+18292 
+5

I don't have a procedure for solving this, I just guessed, had to modify it once, and came up with:

x = -1     y = -3    z = -2

From these values, you can determine the value of x2 + y2 + z2.

 Sep 26, 2015
 #1
avatar+18292 
+5
Best Answer

I don't have a procedure for solving this, I just guessed, had to modify it once, and came up with:

x = -1     y = -3    z = -2

From these values, you can determine the value of x2 + y2 + z2.

geno3141 Sep 26, 2015
 #2
avatar+107359 
+5

x^2 + 6y = -17   →    y = -[17 + x^2] / [6]   (1)

y^2 + 4z = 1  →  [1 - y^2] /[4] = z    (2)

z^2 + 2x = 2   →  x = [2 - z^2] /  [2]     (3)

 

Put (1) into (2)

 

z = [1 - (-[17 + x^2] / [6])^2 ] / 4  simplify

z = [1 - [(289 +34x^2 + x^4)]/36] / 4

z = [36 - 289 -34x^2 -x^4]/36]/4

z = [-253 - 34x^2 -x^4]/ 144

z =   -[(x^4 + 34x^2 +253)/144]     sub this into (3)

 

[-[(x^4 + 34x^2 +253)/144]]^2 + 2x = 2 .....  this would be diificult to solve by hand....using a Computer Algebra System, we have that x = -1    ...so x^2 = 1

 

And y = -[17 +(-1)^2'/[-6]   =  -(18)/6  = -3      ...so y^2  = 9

 

And z = [1 - (-3)^2]/ 4 = (-8)/ 4  = -2    so z^2  = 4

 

So

 

x^2 + y^2 + z^2 =   1 + 9 + 4 = 14

 

 

cool cool cool

 Sep 27, 2015

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