 Mathematics
solve the algebra equation.

### 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 Student has agreed that all tutoring, explanations, and answers provided by the tutor will be used to help in the learning process and in accordance with Studypool's honor code & terms of service. 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) Wallace H (1013)
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