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____ [38]
3 years ago
5

X^2-12x+12=0 completing the square

Mathematics
1 answer:
Makovka662 [10]3 years ago
3 0

Answer:

x1 = 6 + 2√6 or x2 = 6 - 2√6

Step-by-step explanation:

x² - 12x + 12 = 0

We need to use formula:

a² - 2ab + b² = (a - b)².

x² - 12x  =  - 12

x² - 2*6x + 6²- 6² = - 12

(x - 6)² - 36 = -12

(x - 6)² = - 12 + 36

(x -6)² = 24

(x - 6) = √24 or (x-6) = -√24

(x - 6) = 2√6 or (x-6) = - 2√6

x = 6 + 2√6 or x = 6 - 2√6

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2^{51} mod 22 in words, two to the power of fifty-one mod twenty-two
andreyandreev [35.5K]

Since 2⁵ = 32, and

2⁵ ≡ 32 ≡ 10 (mod 22),

we have

2⁵¹ ≡ 2 • 2⁵⁰ ≡ 2 • (2⁵)¹⁰ ≡ 2 • 10¹⁰ (mod 22)

Now consider 10¹⁰ (mod 22):

10 = 2 • 5

10¹⁰ ≡ 2¹⁰ • 5¹⁰ ≡ (2⁵)² • 5¹⁰ ≡ 10² • 5¹⁰ ≡ 2² • 5¹² (mod 22)

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2⁵¹ ≡ 2³ • 5¹² (mod 22)

Now consider 5¹² (mod 22):

5 and 22 are coprime, and ɸ(22) = 10 (where ɸ(<em>n</em>) is the Euler totient function). By Euler's theorem,

5¹² ≡ 5² • 5¹⁰ ≡ 5² • 1 ≡ 25 ≡ 3 (mod 22)

and so

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Another, more tedious method: Start with smaller powers of 2 and look for a pattern.

2 ≡ 2 (mod 22)

2² ≡ 4 (mod 22)

2³ ≡ 8 (mod 22)

2⁴ ≡ 16 (mod 22)

2⁵ ≡ 32 ≡ 10 (mod 22)

2⁶ ≡ 2 • 32 ≡ 2 • 10 ≡ 20 (mod 22)

2⁷ ≡ 2 • 20 ≡ 40 ≡ 18 (mod 22)

2⁸ ≡ 2 • 18 ≡ 36 ≡ 14 (mod 22)

2⁹ ≡ 2 • 14 ≡ 28 ≡ 6 (mod 22)

2¹⁰ ≡ 2 • 6 ≡ 12 (mod 22)

2¹¹ ≡ 2 • 12 ≡ 24 ≡ 2 (mod 22)

2¹² ≡ 2 • 2 ≡ 4 (mod 22)

and so on, with a cyclic pattern of length 10. That is, 2^{10k+1}\equiv2\pmod{22} for any integer <em>k</em> ≥ 0. So 2⁵¹ ≡ 2 (mod 22).

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