SOLUTION: Solve by completing the square x^2-2x-5=0

Question 347531: Solve by completing the square
x^2-2x-5=0
Found 2 solutions by fractalier, unlockmath:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
Okay from
x^2 - 2x - 5 = 0
we group our terms like this
(x^2 - 2x + something) - 5 = 0 + something
The "something" is always half of the middle term's coefficient squared...here it is 1.
so we have
(x^2 - 2x + 1) - 5 = 0 + 1
(x - 1)^2 - 5 = 1
(x - 1)^2 = 6
take square root of both sides now
x - 1 = +/- sqrt(6)
x = 1 +/- sqrt(6)
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
First let's add 6 to both sides of the equal sign to be:
x^2-2x+1=6
You'll see now we can complete the square which means in simple language both factors are the same:
(x-1)(x-1)=6
Or rewritten as:
(x-1)^2= 6
Now to get rid of the 2nd power, we square root both sides and it gives us:
x-1=sq rt 6
Now we add 1 to both sides to get:
x=1+sq rt 6
x=1- sq rt 6
Make sense?
RJ
www.math-unlock.com
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