Using Vieta's formulas, for a cubic equation ax³ + bx² + cx + d = a (x - α) (x - β) (x - γ) = 0,

α + β + γ = -b/a

αβ + αγ + βγ = c/a

αβγ = -d/a


So for √2014 x³ - 4029x² + 2 = 0,

a = √2014, b = -4029, c = 0 , d = 2

x₁ + x₂ + x₃ = 4029/√2014 ①
x₁x₂ + x₁x₃ + x₂x₃ = 0 ②
x₁x₂x₃ = -2/√2014 ③


From ②,

x₁x₂ + x₂x₃ = -x₁x₃
x₂(x₁ + x₃) = -x₁x₃
x₂(x₁ + x₃) = -(-2/√2014)/x₁ (from ③)
x₂(x₁ + x₃) = 2 / √2014 x₁

Next, observe that -4029 = -4028 - 1 = -2(2014) - 1 = -2(√2014)² - 1


√2014 x³ - 4029 x² + 2 = 0 becomes :

(√2014 x)x² - 2(√2014)² x² - x² + 2 = 0

(√2014 x)x² - x² - 2(√2014 x)² + 2 = 0

(√2014 x - 1)x² - 2((√2014 x)² - 1) = 0

(√2014 x - 1)x² - 2(√2014 x - 1)(√2014 x + 1) = 0

(√2014 x - 1)(x² - 2(√2014 x + 1)) = 0

(√2014 x - 1)(x² - 2√2014 x - 2) = 0


√2014 x = 1 or x = [-(-2√2014) ± √((2√2014)² - 4(1)(-2))] / 2(1)


x = 1/√2014 or x = [2√2014 ± √(4(2014 + 2))] / 2

x = √2014 / 2014 or x = √2014 ± √2016


Arranging the roots in order,


x1 = √2014 - √2016
x2 = 1/√2014
x3 = √2014 + √2016



Using the earlier part,


x2(x1 + x3)

= 2 / ( √2014 (1/√2014) )

= 2

Alternatively ,

(√2014 x - 1)(x² - 2√2014 x - 2) = 0

x = 1/√2014 or x² - 2√2014 x - 2 = 0

Using Vieta's formula for the quadratic expression on the right,


sum of roots = -(-2√2014)/1 = 2√2014
product of roots = -2/1 = -2

Since the product of roots is negative, one root must be positive and the other negative.

So, for their sum to be positive, one root must be > 2√2014 (positive) and the other < 0 (negative)


This confirms that x = 1/√2014 is the middle root x2 , since 1/√2014 = √2014 / 2014 and 0 < √2014 / 2014 < 2√2014


So, x1 < 0, x3 > 2√2014, and their sum = 2√2014


Therefore, x2(x1 + x3)

= (1/√2014) (2√2014)

= 2

Assuming that there is a typo error,

x + y + z = 9 ①
xy + yz + zx = 24 ②


From ①,

x + y = 9 - z →③

(x + y)² = (9 - z)² →④


From ②,

xy + z(x + y) = 24
xy + z(9 - z) = 24 (from ③)

xy = 24 - 9z + z² → ⑤


Next,


(x + y)² = x² + 2xy + y²

(x + y)² - 4xy = x² - 2xy - y²

(x + y)² - 4xy = (x - y)²

Sub ③ and ⑤,

(9 - z)² - 4(24 - 9z + z²) = (x - y)²

81 - 18z + z² - 96 + 36z - 4z² = (x - y)²

-3z² + 18z - 15 = (x - y)²


Now, since x , y , z are all real numbers,

(x - y)² ≥ 0 (x - y is also real, and the square of a real number is either 0 or positive)


-3z² + 18z - 15 ≥ 0

3z² - 18z + 15 ≤ 0

z² - 6z + 5 ≤ 0

(z - 1)(z - 5) ≤ 0

Solving by sketching the curve or substitution of values,

1 ≤ z ≤ 5


∴ Greatest value of z = 5

Alternatively

x + y + z = 9 ①
xy + yz + zx = 24 ②

We can let x,y,z be the roots of a cubic equation at³ + bt² + ct + d = 0

Since x,y,z are all real, we have 2 possibilities for the roots :

-All are real and distinct
-2 roots are equal and repeated, the third is different.


(Note that the case of all three roots being equal and repeated is not possible, as that would imply that x = y = z = 3 , since x + y + z = 3z = 3x = 3y = 9

Then, xy + yz + zx = 3² + 3² + 3² = 27 ≠ 24)


Letting t be the variable, rewrite the equation as the following :


(t - x)(t - y)(t - z) = 0

(t - z)(t² - (x+y)t + xy) = 0

Since we want real roots, the quadratic factor must have a discriminant that is 0 or bigger (b² - 4ac ≥ 0)


(x + y)² - 4(1)(xy) ≥ 0

(x + y)² - 4xy ≥ 0

we substitute from ① and ② to reach the same result as before.