Respuesta :

adioabiola

The two numbers which add to 6 and multiply to 58 are; 3+7i and 3-7i

Simultaneous equations;

According to the question;

  • let the two numbers be; x and y.

Hence;

  • x+ y = 6
  • xy = 58

  • x = 6-y which can the be be substituted into equation (ii) so that we have;

(6-y)y = 58

6y -y² = 58

y² - 6y +58 = 0

By solving quadratically; we have;

  • y = 3 + 7i Or y = 3-7i

  • x = 3 - 7i Or x = 3+7i

Read more on complex roots;

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