Answer:
x = - 3[tex]\sqrt{2}[/tex], x = [tex]\sqrt{2}[/tex]
Step-by-step explanation:
Given
x² + 2[tex]\sqrt{2}[/tex] x - 6
To calculate the zeros equate to zero
x² + 2[tex]\sqrt{2}[/tex] x - 6 = 0 ( add 6 to both sides )
x² + 2[tex]\sqrt{2}[/tex] x = 6
One way is to use the completing the square method
add ( half the coefficient of the x- term )² to both sides
x² + 2[tex]\sqrt{2}[/tex] x + ([tex]\sqrt{2}[/tex])² = 6 + ([tex]\sqrt{2}[/tex])²
x² + 2[tex]\sqrt{2}[/tex] x + 2 = 6 + 2
(x + [tex]\sqrt{2}[/tex] )² = 8
Take the square root of both sides
x + [tex]\sqrt{2}[/tex] = ± [tex]\sqrt{8}[/tex] = ± 2[tex]\sqrt{2}[/tex]
subtract [tex]\sqrt{2}[/tex] from both sides
x = - [tex]\sqrt{2}[/tex] ± 2[tex]\sqrt{2}[/tex]
x = - [tex]\sqrt{2}[/tex] - 2[tex]\sqrt{2}[/tex] = - 3[tex]\sqrt{2}[/tex], or
x = - [tex]\sqrt{2}[/tex] + 2[tex]\sqrt{2}[/tex] = [tex]\sqrt{2}[/tex]
