3.0k views
1 vote
Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.

u =

x

4(x +1)
x +1

(x + 1)2
Select the solution(s) of the original equation.

x = 1 + sqrt 2

x = 2 + sqrt 2

x = 3 + sqrt 2
x = 1 - sqrt 2

x = 2 - sqrt 2

x = 3 - sqrt 2

User Nidhi
by
6.2k points

2 Answers

4 votes

Answer: 1st - x = 1 + √2 & 4th- x= 1 - √2

Explanation:

Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution. u = x 4(x +1) x +1 (x + 1)2 Select-example-1
User Gbs
by
6.1k points
5 votes

For this case we have to:

Let
u = x + 1

So:


u ^ 2-4u + 2 = 0

We have the solution will be given by:


u = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Where:


a = 1\\b = -4\\c = 2

Substituting:


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}

The solutions are:


u_ {1} = \frac {4 + 2 \sqrt {2}} {2} = 2 + \sqrt {2}\\u_ {2} = \frac {4-2 \sqrt {2}} {2} = 2- \sqrt {2}

Returning the change:


2+ \sqrt {2} = x_ {1} +1\\x_ {1} = 1 + \sqrt {2}\\2- \sqrt {2} = x_ {2} +1\\x_ {2} = 1- \sqrt {2}

Answer:


x_ {1} = 1 + \sqrt {2}\\x_ {2} = 1- \sqrt {2}

User Limpuls
by
6.5k points