f5數學題一問?10點

1.
Let y = x + (2/x) .Express x² + (4/x²) in terms of y

Solution :
x + (2/x) = y
[x + (2/x)]² = y²
x² + 2x(2/x) + (2/x)² = y²
x² + 4 + (4/x²) = y²
x² + (4/x²) = y² - 4


2.
Solve equation x² + 3x + 4 + (6/x) + (4/x²) = 0,and express youranswers in form a + bi

x² + 3x + 4 + (6/x) + (4/x²) = 0
x² + (4/x²) + 3x + (6/x) + 4 = 0
[x² + (4/x²)] + 3[x + (2/x)] + 4 = 0

From Q.1, when x + (2/x) = y, then x² + (4/x²) = y².
(y² - 4) + 3y + 4 = 0
y² + 3y = 0
y(y + 3) = 0
y = 0 or y = -3
x + (2/x) = 0 or x + (2/x) = -3
(x² + 2)/x = 0 or (x² + 2)/x = -3
x² + 2 = 0 or x² + 2 = -3x
x² = -2 or x² + 3x + 2 = 0
x² = -2 or (x + 1)(x + 2) = 0
x = (√2)i or x = -(√2)i or x = -1 or x = -2
x = 0 + (√2)i or x = 0 - (√2)i or x = -1 + 0i or x = -2 + 0i

1.y = x + 2/x
y^2 = (x + 2/x)^2 = x^2 + 4 + 4/x^2
so x^2 + 4/x^2 = y^2 - 4.
2.
x^2 + 4/x^2 + 3(x + 2/x) + 4 = 0
Using the above result,
y^2 - 4 + 3y + 4 = 0
y(y + 3) = 0
y = 0 or - 3
that is
x + 2/x = 0 (reject, no real roots) or x + 2/x = - 3.
x^2 + 3x + 2 = 0
(x + 2)(x + 1) = 0
so x = - 2 or - 1.