MATH SOLVE

4 months ago

Q:
# 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 )

Accepted Solution

A:

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)

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)