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Zolol [24]
3 years ago
15

Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.

Mathematics
2 answers:
lara31 [8.8K]3 years ago
5 0

Answer: c) x+1

Step-by-step explanation:

trust me on edg

solniwko [45]3 years ago
3 0

For this case we have to:

Letu = 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}

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Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

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Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

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the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

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Step-by-step explanation:

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