Answer:
x = - 4 ± 2[tex]\sqrt{3}[/tex]
Step-by-step explanation:
Given
f(x) = x² + 8x + 4
To find the zeros let f(x) = 0, that is
x² + 8x + 4 = 0 ( subtract 4 from both sides )
x² + 8x = - 4
To solve using the method of completing the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(4)x + 16 = - 4 + 16
(x + 4)² = 12 ( take the square root of both sides )
x + 4 = ± [tex]\sqrt{12}[/tex] = ± 2[tex]\sqrt{3}[/tex] ( subtract 4 from both sides )
x = - 4 ± 2[tex]\sqrt{3}[/tex]
Thus the zeros are
x = - 4 - 2[tex]\sqrt{3}[/tex] and x = - 4 + 2[tex]\sqrt{3}[/tex]
