Respuesta :
Answer:
Step-by-step explanation:
We are given the equation [tex](x+1)^2-4(x+1)+2=0[/tex]. Consider the substitution u = (x+1). Then the equation turns out to be [tex]u^2-4u+2=0[/tex]
Recall that given a second degree polynomial of the form [tex]ax^2+bx+c =0[/tex] then its solutions are given by the expression
[tex] x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}[/tex]
In our case, a = 1, b = -4 and c =2. Then the solutions are
[tex]u_1 = \frac{4+\sqrt[]{(-4)^2-4(2)}}{2} = 2 +\sqrt[]{2}[/tex]
[tex]u_2 = \frac{4-\sqrt[]{(-4)^2-4(2)}}{2} = 2 -\sqrt[]{2}[/tex]
We have that x = u-1. So the original solutions are
[tex]x_1 = 2 +\sqrt[]{2}-1 = 1 + \sqrt[]{2}[/tex]
[tex]x_2 =2 -\sqrt[]{2} -1 = 1 - \sqrt[]{2}[/tex]
Answer: x+4
Step-by-step explanation: part one- x+4
Part two- 3+sqrt21 over 2 +4 & 3- sqrt21 over 2 +4