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

kes1968 Sep 26, 2015

#1**+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 x^{2} + y^{2} + z^{2}.

geno3141 Sep 26, 2015

#2**+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

CPhill Sep 27, 2015