SOLUTION: Help me please. It asks to solve by completing the square. x^2-x-1=0 What I got for that was (y=a) (a=1) (H=1/2) and (k=5/4) I am not sure if they are correct or not, please hel

Algebra ->  Algebra  -> Radicals -> SOLUTION: Help me please. It asks to solve by completing the square. x^2-x-1=0 What I got for that was (y=a) (a=1) (H=1/2) and (k=5/4) I am not sure if they are correct or not, please hel      Log On

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Question 136355: Help me please. It asks to solve by completing the square.
x^2-x-1=0
What I got for that was (y=a) (a=1) (H=1/2) and (k=5/4)
I am not sure if they are correct or not, please help,Thank You=)
Answer by Edwin McCravy(6935) About Me  (Show Source):
You can put this solution on YOUR website!
Help me please. It asks to solve by completing the square.
x^2-x-1=0
What I got for that was (y=a) (a=1) (H=1/2) and (k=5/4)
I am not sure if they are correct or not, please help,Thank You=)

x%5E2-x-1=0
Get rid of the constant term -1 on the left by adding %2B1
to both sides:
x%5E2-x=1

Multiply the coefficient of x on the left, which is -1 by 1%2F2, getting -1%2F2.
Now square -1%2F2. %28-1%2F2%29%5E2=1%2F4
Now add 1%2F4 to both sides:
x%5E2-x%2B1%2F4=1%2B+1%2F4
write the 1 as 1%2F1
x%5E2-x%2B1%2F4=1%2F1%2B+1%2F4
The LCD on the right is 4, so multiply 1%2F1 by 4%2F4
x%5E2-x%2B1%2F4=+++%281%2F1+%29%2A%284%2F4%29%2B+1%2F4++
x%5E2-x%2B1%2F4=4%2F4%2B+1%2F4
x%5E2-x%2B1%2F4=5%2F4
Now factor the left side:
%28x-1%2F2%29%28x-1%2F2%29=5%2F4
The left side is a perfect square (that's why the method is
called "completing the square"). So we write the left side
as a square, the square of a binomial:
%28x-1%2F2%29%5E2=5%2F4
Next we use the principle of square roots:
x-1%2F2= ±sqrt%285%2F4%29
x-1%2F2= ±sqrt%285%29%2Fsqrt%284%29
x-1%2F2= ±sqrt%285%29%2F2
Now we solve for x by adding 1%2F2 to both sides:
x-1%2F2%2B1%2F2= 1%2F2%2B-sqrt%285%29%2F2
x= 1%2F2%2B-sqrt%285%29%2F2
Edwin