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lidiya [134]
3 years ago
7

What are the solutions of x^2-2x+26=0?

Mathematics
2 answers:
Kazeer [188]3 years ago
8 0

Answer:

x = 1 + (i)5 and x = 1 - (i)5

Step-by-step explanation:

x^2-2x+26=0 can be rewritten by completing the square of x^2-2x, as follows:

x^2-2x+26=0

x^2-2x+ 1  - 1  +26=0 (this 1 comes from halving the coefficient of x (which is -2), obtaining -1, squaring the result, and then adding this 1 to and subtracting this 1 from x^2-2x)  →  x^2-2x+ 1  - 1  +26=0

Rewriting x^2-2x+ 1 as the square of a binomial, we get:

 (x - 1)²  - 1  +26=0, or  (x - 1)²  - 1  +26=0, or   (x - 1)² = -25

Taking the square root of both sides yields x - 1 = ±(i)5.

Thus, the roots are x = 1 + (i)5 and x = 1 - (i)5  (Answer C)

zhuklara [117]3 years ago
7 0

I think it's 2+4×5=6 or 7+910=10

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List all subsets of ta, b, c, d, e) containing a but not containing b
fomenos

Answer:

(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)

Step-by-step explanation:

We are given the set (a,b,c,d,e).

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Subsets:

φ

(a,b,c,d,e)

(a), (b), (c), (d), (e)

(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e)

(a,b,c), (a,b,d), (a,b,e), (a,c,d), (a,c,e), (a,d,e), ( b,c,d), (b,c,e), (b,d,e), (c,d,e)

(a,b,c,d), (a,b,c,e), (a,b,d,e), (a,c,d,e), (b,c,d,e)

Subset having a but not b :

(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)

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