What is the difference of the factors of the expression x2 + (1/x2) – 6?
Answers
x² + 1/x² - 6 = 0
x² + 1/x² - 2 - 4 = 0
x² + 1/x² - 2(x* 1/x) - 4 = 0
(x - 1/x)² - 2² = 0
(x - 1/x + 2) (x - 1/x - 2)
Their difference is 4
Answer:
4 & 4√2
x = 1 ± √2 or x = - 1 ± √2
Δx = 2 or 2√2
Step-by-step explanation:
x² + 1/x² = 6
=> x² + 1/x² - 6 = 0
=> x² + 1/x² + 2 - 8 = 0
=> x² + 1/x² + 2x(1/x) - (2√2)² = 0
=> (x + 1/x)² - (2√2)² = 0
=> (x + 1/x + 2√2)(x + 1/x - 2√2) = 0
(x + 1/x + 2√2) - (x + 1/x - 2√2) = 4√2
x² + 1/x² = 6
=> x² + 1/x² - 6 = 0
=> x² + 1/x² - 2 - 4 = 0
=> x² + 1/x² - 2x(1/x) - (2)² = 0
=> (x - 1/x)² - (2√2)² = 0
=> (x - 1/x + 2)(x - 1/x - 2) = 0
(x - 1/x + 2) - (x - 1/x - 2) = 4
if solving for x
x² + 1/x² - 6 = 0
=> x² + 1/x² - 2 - 4 = 0
=> ( x - 1/x)² = 2²
=> (x - 1/x) = ±2
=> x² - 1 = ±2x
=> x² - 2x - 1 = 0 or x²+2x - 1 = 0
=> x = (2 ± √4 + 4)/2 or x = -2 ±√4 + 4)/2
=> x = 1 ± √2 or x = - 1 ± √2
Δx = 2 or 2√2