SOLUTION: 2x^2-3x+1=0 Solve by completing the square.

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Question 166950: 2x^2-3x+1=0
Solve by completing the square.
Found 2 solutions by nerdybill, Edwin McCravy:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2-3x+1=0
2x^2-3x = -1
2(x^2 - (3/2)x) = -1
2(x^2 - (3/2)x + 9/16) = -1 + 9/8
2(x-(3/4))^2 = -8/8 + 9/8
2(x-(3/4))^2 = 1/8
(x-(3/4))^2 = 1/16
x-(3/4) = (+-)1/4
x = (3/4)(+-)1/4
.
x = {2/4, 4/4}
x = {1/2, 1}

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Edwin's solution:


2x%5E2-3x%2B1=0

Isolate the variable terms by
adding -1 to both sides:

2x%5E2-3x=-1

Divide thru by 2, the coefficient of the 
term in x%5E2

%282%2F2%29x%5E2-%283%2F2%29x=-1%2F2

x%5E2-%283%2F2%29x=-1%2F2

To the side, multiply the coefficient of x by 1%2F2.
-3%2F2%29%281%2F2%29=-3%2F4

Now square -3%2F4.  %28-3%2F4%29%5E2=9%2F16

Add 9%2F16 to both sides of

x%5E2-%283%2F2%29x%2B9%2F16=-1%2F2%2B9%2F16

Combine the right side: -1%2F2%2B9%2F16=-8%2F16%2B9%2F16=1%2F16

x%5E2-%283%2F2%29x%2B9%2F16=1%2F16

Factor the left side:

%28x-3%2F4%29%28x-3%2F4%29=1%2F16

Write the left side as the square of
a binomial:

%28x-3%2F4%29%5E2=1%2F16

Take the square root of both sides,
remembering the ± on the right:

sqrt%28%28x-3%2F4%29%5E2%29+=+%22%22+%2B-sqrt%281%2F16%29

x-3%2F4=±1%2F4

x=3%2F4%2B-1%2F4

Using the +

x=3%2F4%2B1%2F4
x=4%2F4
x=1

Using the -

x=3%2F4-1%2F4
x=2%2F4
x=1%2F2

The solutions are 1 and 1%2F2

Edwin