1. ## Completing the square

Can anyone solve this quadratic by completing the square for me?
$\displaystyle 16y^2+16y=1$
Steps you took?

2. ## Solution

It's been a few years since I've had a problem like this, so hopefully I am going about this correctly. Start by dividing the equation through by 16 to get:

y^2 + y = 1/16

Now, we figure that in order to factor the left hand side of the equation we would need something of the form (y + c)^2 with c denoting some constant value.

We would like to choose something along the lines of:

(y+1/2)^2 because when you multiply it out you get:

y^2 + y + 1/4. But since we have this additional 1/4 on the left side of the equation, we must add it to the right side to get the full equation:

y^2 + y + 1/4 = 1/16 + 1/4

This becomes: (y + 1/2)^2 = (1/4) + (1/16)
which becomes: (y + 1/2)^2 = 5/16.

Now take the square root of both sides to yield:

y + 1/2 = +/- sqrt(5)/4

Final step: subtract 1/2 from both sides to solve for y:

y = (+/- sqrt(5)/4) - 1/2

I hope this explanation is useful.

3. Most. Thanx a bunch.

4. Originally Posted by EyesForEars
Can anyone solve this quadratic by completeing the square for me?
$\displaystyle 16y^2+16y=1$
Steps you took?
Edit: This is late, but here it is anyway.

First thing, we have to make the coefficient of the quadratic term 1. So divide all terms by 16.

$\displaystyle 16y^2+16y=1$

$\displaystyle y^2+y=\frac{1}{16}$

Now, take half of the coefficient of the linear term, square it, and add it to both sides.

$\displaystyle y^2+y+\frac{1}{4}=\frac{1}{16}+\frac{1}{4}$

$\displaystyle (y+\frac{1}{4})^2=\frac{5}{16}$

Now, take the square root of both sides.

$\displaystyle y+\frac{1}{2}=\frac{\pm \sqrt{5}}{4}$

$\displaystyle y=\frac{\pm \sqrt{5}}{4}-\frac{1}{2}=\frac{\pm \sqrt{5}-2}{4}$