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What are the solutions to the equation over the complex numbers?

Cant write as i dont understand the formatting

What are the solutions to the equation over the complex numbers? Cant write as i dont-example-1

2 Answers

7 votes


\large\huge\green{\sf{Question:-}}

  • ➡x²+50=0


\large\huge\green{\sf{Answer:-}}

we know

if x²= y then x = ±√y


x {}^(2) + 50 = 0 \\ {x}^(2) = ( - 50) \\ x = ±√(( - 50)) \\ x =± √(50 *( - 1)) \\ ➡x = ±√((2 * 5 * 5) * i) \\ ➡x = ±5 √(2i) \\ ➡x = + 5 √(2i) and \: - 5 √(2i)

User Tomer Zeitune
by
3.8k points
5 votes

Answer:

4i=4 cis (pi/2).

Explanation:

In relation to quadratic equations, imaginary numbers (and complex roots) occur when the value under the radical portion of the quadratic formula is negative. When this occurs, the equation has no roots (or zeros) in the set of real numbers.

User DhruvPathak
by
3.4k points