SOLUTION: solve this equation by completing the square: 1x^2/4+ 1x/16+ 1/8= 0

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Question 248605: solve this equation by completing the square:
1x^2/4+ 1x/16+ 1/8= 0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2F4%29x%5E2%2B+%281%2F16%29x%2B+1%2F8=+0
To complete the square:
  1. Transform the equation so that the variable terms are on one side and the constant term is on the other side.
  2. If the leading coefficient is:
    • a 1:
      1. Calculate 1/2 of the coefficient of x. Let's call this "q". (Don't forget to keep track of the sign of q.)
      2. Add q%5E2 to both sides of the equation
    • Not a 1:
      1. Factor out the leading coefficient. (This can require some creative factoring.)
      2. Calculate 1/2 of the coefficient of x (inside the factor). Let's call this "q". (Don't forget to keep track of the sign of q.)
      3. Add q%5E2 inside the factor with the variables and add (leading coefficient)*q%5E2 to the other side of the equation.(This may seem that you are adding different things to each side. But with the leading coefficient in front of the factor on the left side, you are really adding (leading coefficient)*q%5E2 to both sides.)
  3. Factor the trinomial (3-term expression) you have just created into %28x+%2B+q%29%5E2

Since completing the square is easier if the leading coefficient is 1, I will start by multiplying both sides by 4:
x%5E2%2B+%281%2F4%29x%2B+1%2F2=+0
(Note: When you are working with ellipses and hyperbolas (which are certain two variable equations), you will not be able to force a leading coefficient of 1 like we have here.)

Now that we have an easier equation to work with, let's go through the procedure:
1. Transform the equation...
Subtract 1/2 from each side:
x%5E2%2B+%281%2F4%29x+=+%28-1%29%2F2
2. Since the leading coefficient is 1, calculate "q" (1/2 of the coefficient of x):
%281%2F2%29%281%2F4%29+=+1%2F8
3. Add to both sides. (Since our leading coefficient is 1, we can just add q%5E2. q%5E2+=+%281%2F8%29%5E2+=+1%2F64
x%5E2%2B+%281%2F4%29x+%2B+1%2F64+=+%28-1%29%2F2+%2B+1%2F64
x%5E2%2B+%281%2F4%29x+%2B+1%2F64+=+%28-32%29%2F64+%2B+1%2F64
x%5E2%2B+%281%2F4%29x+%2B+1%2F64+=+%28-31%29%2F64
4. Factor the trinomial into %28x%2Bq%29%5E2
%28x%2B+1%2F8%29%5E2+=+%28-31%29%2F64

Now that we have completed the square we are in position to solve the equation. Normally the next step would be to find the square root of both sides of the equation. Then we would solve the two equations that result.

However this equation says that %28x%2B1%2F8%29%5E2 is a negative number. There are no real numbers whose square is negative. So there are no solutions (that are Real numbers).

If you want complex solutions then we will proceed:
sqrt%28%28x%2B+1%2F8%29%5E2%29+=+sqrt%28%28-31%29%2F64%29
abs%28x%2B+1%2F8%29+=+%28sqrt%2831%29%2F8%29i
x%2B+1%2F8+=+%28sqrt%2831%29%2F8%29i or x%2B+1%2F8+=+%28-sqrt%2831%29%2F8%29i%29
Add %28-1%29%2F8} to (or subtract 1/8 from) each side:
x+=+%28-1%29%2F8+%2B+%28sqrt%2831%29%2F8%29i or x+=+%28-1%29%2F8+%2B+%28-sqrt%2831%29%2F8%29i

In case you're interested, here is what completing the square looks like when the leading coefficient is not 1:
%281%2F4%29x%5E2%2B+%281%2F16%29x%2B+1%2F8=+0
1. Transform the equation.
Subtract 1/8:
%281%2F4%29x%5E2%2B+%281%2F16%29x+=+%28-1%29%2F8
2. Since the leading coefficient is not 1,
2a. Factor out the leading coefficient. (This is where some creativity may be needed.):
%281%2F4%29%28x%5E2%2B+%281%2F4%29x%29+=+%28-1%29%2F8
2b. Calculate "q"
q+=+%281%2F2%29%281%2F4%29+=+1%2F8
2c. Add q%5E2 in side the parentheses and (leading coefficient)*q%5E2 to the right side. q%5E2+=+%281%2F8%29%5E2+=+1%2F64:
%281%2F4%29%28x%5E2%2B+%281%2F4%29x+%2B+1%2F64%29+=+%28-1%29%2F8+%2B+%281%2F4%29%281%2F64%29
which simplifies to:
%281%2F4%29%28x%5E2%2B+%281%2F4%29x+%2B+1%2F64%29+=+%28-1%29%2F8+%2B+1%2F256
%281%2F4%29%28x%5E2%2B+%281%2F4%29x+%2B+1%2F64%29+=+%28-32%29%2F256+%2B+1%2F256
%281%2F4%29%28x%5E2%2B+%281%2F4%29x+%2B+1%2F64%29+=+%28-31%29%2F256
3. Rewrite as a perfect square:
%281%2F4%29%28x%2B1%2F8%29%5E2+=+%28-31%29%2F256
If we multiply both sides of this equation by 4, we end up with the same equation as before:
%28x%2B+1%2F8%29%5E2+=+%28-31%29%2F64