Question

What are the roots of the equation x²+2x+17=0 in simplest a+bi form?​

Answer 1

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The required roots of given equation are ~

• $\boxed{ \boxed{ - 1 + 4i} \: \: and \: \: \boxed{ - 1 - 4i}}$

The solution is in attachment ~

Answer 2

Answer:

  -1 ±4i

Step-by-step explanation:

The quadratic can be written in vertex form as ...

  x² +2x +17 = 0

  (x² +2x +1) +16 = 0 . . . . . . . . identify the part that can be a perfect square

  (x +1)² = -16 . . . . . . . . . . . . subtract 16

  x +1 = ±√(-16) = ±4i . . . . take the square root

  x = -1 ±4i . . . . . . . . . . . subtract 1

The roots of the equation are x = -1-4i and x = -1+4i.

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Additional comment

It can be worthwhile to remember the form of a perfect square trinomial:

  (x +a)² = x² +2ax +a²

If you know the coefficient of x, you can use half of it as the constant in each binomial factor.

In the above, we had 2a=2, so a=1 and a²=1. The constant 17 then is resolved into two parts: one part (1) is used to "complete the square" and the other part (16) is the y-coordinate of the vertex. It is used to find the "radical" portion of the roots.