Solve each equation by completing the square using the steps in the lesson content.?

Question

x2 + 2x + 5 = 0 ... I got x = -1 and x = -1 - 2i , but don't know if that is right.

Answer 1

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Answer: x = -1 + 2i and x = -1 - 2i

x² + 2x + 5 = 0

Second term is the key here.

Remember

(a + b)² = a² + 2ab + b²

and recognize that in your equation 2x can be written as 2*x*1 so a=x and b=1

(x + 1)² = x² + 2x + 1

What is needed to make this equal to the left side of our equation? Answer: we have to add 4.

 x² + 2x + 5 = (x² + 2x + 1) + 4 =

= (x + 1)² + 4 = 0

from here

(x + 1)² = -4

square root of both sides

x + 1 = ±√(-4)

we'll find this root step by step to make it clear

x + 1 = ±√(4 * (-1)) = ±2√(-1) = ±2i

subtract 1 from both sides

x = -1 ± 2i

Our two solutions are

x = -1 + 2i

x = -1 - 2i

Answer 2

x² + 2x + 5 => (x + 1)² + 5 - 1

i.e. (x + 1)² + 4 = 0

so, (x + 1)² = -4

Then, x + 1 = ±2i

Hence, x = -1 ± 2i

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Answer 3

x^2 + 2x + 5 = 0

(x + 1)^2 + 4 = 0

Complex solutions:

x = -1 - 2 i

x = -1 + 2 i

Answer 4

No.  It's not correct.

You'd either get 0 complex roots or 2 complex roots.  You would never only get one.

Let's see what I get:

x² + 2x + 5 = 0

Let's start with moving the 5 to the other side:

x² + 2x = -5

Now complete the square by adding 1 to both sides (square of half of 2):

x² + 2x + 1 = -4

Now we can factor the left side:

(x + 1)² = -4

Square root of both sides:

x + 1 = ±2i

subtract 1 from both sides:

x = -1 ± 2i

One of your roots were correct, but both roots are complex numbers.

complex and irrational roots in a polynomial are always in conjugate pairs.