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Find the solution set of the quadratic equation over the set of complex numbers.

2x2 – 6x + 7 = 0














A)


x =
3
2
i or 2i


B)









x =
1
7
(3 − i
5
) or
1
7
(3 + i
5
)


C)









x =
1
2
(3 − i
5
) or
1
2
(3 + i
5
)
Eliminate

D)









x =
1
7
(3 − i
41
) or
1
7
(3 + i
41
)

Respuesta :

carlosego
We have the following equation:
 2x2 - 6x + 7 = 0
 Using the resolver we have:
 x = (- b +/- root (b ^ 2 - 4 * a * c)) / (2 * a)
 Substituting values we have:
 x = (- (- 6) +/- root ((- 6) ^ 2 - 4 * 2 * 7)) / (2 * 2)
 Rewriting we have:
 x = (6 +/- root (36 - 56)) / (4)
 x = (6 +/- root (-20)) / (4)
 x = (6 +/- root (-4 * 5)) / (4)
 x = (6 +/- 2raiz (-5)) / (4)
 x = (6 +/- 2raiz (-1 * 5)) / (4)
 x = (6 +/- 2raiz (5) * i) / (4)
 x = (3 +/- root (5) * i) / (2)
 The solutions are:
 x1 = (3 + root (5) * i) / (2)
 x2 = (3 - root (5) * i) / (2)
 Answer:
 x1 = (3 + root (5) * i) / (2)
 x2 = (3 - root (5) * i) / (2)

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