Answer:
the first one and the fith one
Step-by-step explanation:
Answer:
x = 4 ± 
Step-by-step explanation:
Given
x² - 8x = 3
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(- 4)x + 16 = 3 + 16
(x - 4)² = 19 ( take the square root of both sides )
x - 4 = ±
( add 4 to both sides )
x = 4 ±
Use the chain rule to compute the second derivative:

The first derivative is


Then the second derivative is


Then plug in π/4 for <em>x</em> :
