Ordered triple

Find an ordered triple (x,y,z) of real numbers satisfying x<= y<= z and the system of equations
sqrtx + sqrty +sqrtz = 10
x+y+z=38
(sqrtxyz) = 30

\(\begin{array}{|lrcll|} \hline (1) & \sqrt{x} + \sqrt{y} +\sqrt{z} &=& 10 \\ (2) & x+y+z &=& 38 \\ (3) & \sqrt{xyz} &=& 30 \\ \hline \end{array}\)

1.

\(\begin{array}{|lrcll|} \hline & (\sqrt{x} + \sqrt{y} +\sqrt{z})^2 &=& 10^2 \\ & \underbrace{x+y+z}_{=38} +2( \sqrt{xy}+\sqrt{xz}+\sqrt{yz} ) &=& 100 \\ & 38 +2( \sqrt{xy}+\sqrt{xz}+\sqrt{yz} ) &=& 100 \quad & | \quad -38 \\ & 2( \sqrt{xy}+\sqrt{xz}+\sqrt{yz} ) &=& 62 \quad & | \quad :2 \\ \mathbf{(4)}& \mathbf{\sqrt{xy}+\sqrt{xz}+\sqrt{yz}} &\mathbf{=}& \mathbf{31} \\ \hline \end{array}\)

\(\begin{array}{lcll} \text{We substitute:} \\ \quad x_1 = \sqrt{x} & \text{ or }& x = x_1^2 \\ \quad x_2 = \sqrt{y} & \text{ or }& y = x_2^2 \\ \quad x_3 = \sqrt{z} & \text{ or }& z = x_3^2 \\ \end{array}\)

\(\begin{array}{|rcll|} \hline x_1+x_2+x_3 &=& \sqrt{x} + \sqrt{y} +\sqrt{z} \\ &=& 10 \\ x_1\cdot x_2 \cdot x_3 &=& \sqrt{xyz} \\ &=& 30 \\ x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 &=& \sqrt{xy}+\sqrt{xz}+\sqrt{yz} \\ &=& 31 \\ \hline \end{array}\)

\(\begin{array}{lcll} \text{We set:} \\ \quad p &=& -(x_1+x_2+x_3) \\ &=& -10 \\ \quad q &=& x_1\cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 \\ &=& 31 \\ \quad r &=& -(x_1\cdot x_2 \cdot x_3) \\ &=& -30 \\ \end{array}\)

\(\begin{array}{lrcll} \text{The cubic equation:} \\ \boxed{ x^3 - 10x^2+31x-30 = 0 } \\ \end{array}\)

The first solution is \(x_1 = 2\)

Long division:

\(\begin{array}{rcll} x^2-8x+15 = (x-3)(x-5) \end{array}\)

so \(x_2 = 3\) and \(x_3 = 5\)

\(\begin{array}{|rcll|} \hline x_1 = 2 & x &=& x_1^2 \\ & \mathbf{x} &\mathbf{=}& \mathbf{4} \\\\ x_2 = 3 & y &=& x_2^2 \\ & \mathbf{y} &\mathbf{=}& \mathbf{9} \\\\ x_3 = 5 & z &=& x_3^2 \\ & \mathbf{z} &\mathbf{=}& \mathbf{25} \\ \hline \end{array}\)

The ordered triple (4, 9, 25) of real numbers satisfying \(4\le 9\le 25\) and the system of equations.