Can anyone solve this quadratic by completing the square for me?
$\displaystyle
16y^2+16y=1
$
Steps you took?
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.
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.
$\displaystyle y+\frac{1}{2}=\frac{\pm \sqrt{5}}{4}$
$\displaystyle y=\frac{\pm \sqrt{5}}{4}-\frac{1}{2}=\frac{\pm \sqrt{5}-2}{4}$