solve the algebra equation.
Algebra

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solve the equation by completing the square.
x^2 + 2x = 5
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 xterm (regular x, not x^2).
x^2 + 2x = 5
The xterm 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 <xexpression 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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