SOLUTION: Hey, I'm on a question where you have to solve the question that a student has attempted but is wrong, I can't seem to find the problem. Are you able to help me please? Here

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Question 875488: Hey,
I'm on a question where you have to solve the question that a student has attempted but is wrong, I can't seem to find the problem. Are you able to help me please?
Here is a picture of the students workout and question.
http://gyazo.com/df24ccae44177730e6f7a41db326a995
Be much help, thank you.
Found 2 solutions by rothauserc, Theo:
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
This is a complete the square problem
2x^2 +4x - 4 = 0
divide by 2
x^2 + 2x - 2 = 0
now add -2 to both sides of =
x^2 +2x = 2
take 1/2 of 2 and square it, then add to both sides of =
x^2 +2x + 1 = 3
(x +1)^2 = 3
take square root of both sides of =
x+1 = square root(3)
x = square root(3) - 1
x = +square root(3) - 1
x = -square root(3) -1
To the student, remember when completing the square to move the loose number (-2) to the right of the equal sign and square root means + or - square root

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
here's the link that you showed me.
http://gyazo.com/df24ccae44177730e6f7a41db326a995
here's the corrected solution:

the standard form of the quadratic equation is ax^2 + bx + c
if you divide by a you get:
x^2 + (b/a)x + c/a = 0
you have to take 1/2 of the coefficient of the x term.
you will then get:
(x + b/2a)^2 - (b/2a)^2 + c = 0
the student didn't take half of the coefficient of the x term.
that was the student's first mistake.
then when the student took the square root of both sides of the equation, the student only took the positive side of the square root and not the negative side of the square root.
that was the second mistake.