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zavuch27 [327]
3 years ago
5

Solve for x Xsquare+ 4x+45=0

Mathematics
1 answer:
I am Lyosha [343]3 years ago
6 0

For this case we must solve the following quadratic equation:

x ^ 2 + 4x + 45 = 0

Where:

a = 1\\b = 4\\c = 45

The solution will be given by:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

Substituting the values we have:

x = \frac {-4 \pm \sqrt {4 ^ 2-4 (1) (45)}} {2 (1)}\\x = \frac {-4 \pm \sqrt {16-180}} {2}\\x = \frac {-4 \pm \sqrt {-164}} {2}

By definition we have to:

i ^ 2 = -1

So:

x = \frac {-4 \pm \sqrt {164i ^ 2}} {2}\\x = \frac {-4 \pm i \sqrt {164}} {2}\\x = \frac {-4 \pm i \sqrt {2 ^ 2 * 41}} {2}\\x = \frac {-4 \pm2i \sqrt {41}} {2}\\x = -2 \pm i \sqrt {41}

Thus, we have two complex roots:

x_ {1} = - 2 + i \sqrt {41}\\x_ {2} = - 2-i \sqrt {41}

Answer:

x_ {1} = - 2 + i \sqrt {41}\\x_ {2} = - 2-i \sqrt {41}

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If U = the set of odd natural numbers and B = {13, 15, 17, 19, 21, 23, . . .}, find B′.
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