### Question Description

I’m trying to learn for my Algebra class and I’m stuck. Can you help?

solve the equation by completing the square.

x^2 + 2x = 5

## Final Answer

x^2 + 2x = 5

*First, before we can complete the square, we need to make sure that the x^2 and x terms are on the same side in order respectively, and the constant number (the number with no variable next to it) is on the other side of the equal sign.*

x^2 + 2x = 5 <On the left side the x^2 term comes first and then the x term. And the constant number is

on the right side. Everything is ready for completing the square>

*Next, we will let the variable 'b' represent the coefficient of x-term (regular x, not x^2).*

x^2 + 2x = 5

The x-term is 2x, and its coefficient is 2. Therefore b = 2

*Third, find the value of (b/2)^2*

(b/2)^2 = (2/2)^2 <Since b = 2, we substitute b with 2>

1^2 <Evaluate inside the parenthesis first>

1

*Now, we are ready to complete the square to solve the equation. The process is done as follows:*

x^2 + 2x = 5 <Given equation>

x^2 + 2x + 1 = 5 + 1 <Add both sides by your value of (b/2)^2 >

x^2 + 2x + 1 = 6 <Evaluate the right side>

(x+1)(x+1) = 6 <The left side should represent a perfect square, so we can factor this side>

(x+1)^2 = 6 <x-expression must be in the form (a + b)^2>

sqrt[(x+1)^2] = sqrt(6) <Take the square root of both sides>

|x+1| = sqrt(6) <Taking the square root of a squared expression cancels the exponent and the

inside expression is put in an absolute value expression>

Keep in mind that if you have an absolute value expression set equal to a positive, we can solve for two cases. A positive case, and a negative case.

|x+1| = sqrt(6)

x + 1 = sqrt(6) or x + 1 = -sqrt(6) <Two cases to solve for>

x + 1 - 1 = sqrt(6) - 1 or x + 1 - 1 = -sqrt(6) - 1 <In both cases subtract both sides by 1>

x = sqrt(6) - 1 or x = -sqrt(6) - 1

x = -1 + sqrt(6) or x = -1 - sqrt(6)

**SOLUTION: x = -1 + sqrt(6) or x = -1 - sqrt(6)**

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