For this case we have to:
Let[tex]u = x + 1[/tex]
So:
[tex]u ^ 2-4u + 2 = 0[/tex]
We have the solution will be given by:
[tex]u = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
Where:
[tex]a = 1\\b = -4\\c = 2[/tex]
Substituting:
[tex]u = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (1) (2)}} {2 (1)}\\u = \frac {4 \pm \sqrt {16-8}} {2}\\u = \frac {4 \pm \sqrt {8}} {2}\\u = \frac {4 \pm \sqrt {2 ^ 2 * 2}} {2}\\u = \frac {4 \pm2 \sqrt {2}} {2}[/tex]
The solutions are:
[tex]u_ {1} = \frac {4 + 2 \sqrt {2}} {2} = 2 + \sqrt {2}\\u_ {2} = \frac {4-2 \sqrt {2}} {2} = 2- \sqrt {2}[/tex]
Returning the change:
[tex]2+ \sqrt {2} = x_ {1} +1\\x_ {1} = 1 + \sqrt {2}\\2- \sqrt {2} = x_ {2} +1\\x_ {2} = 1- \sqrt {2}[/tex]
Answer:
[tex]x_ {1} = 1 + \sqrt {2}\\x_ {2} = 1- \sqrt {2}[/tex]
Answer: c) x+1
Step-by-step explanation:
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