1. y^{2}2py15p^{2}=0^{
}
y^{2}2py+ =15p^{2}: Then, Divide the coefficient of y in the middle term (2p) by 2 (that will be 1p) and square it. You get 1p^{2} or p^{2} (they are the same). That is completing the square.
p^{2} is your new third term.
y22py+p^{2}=15p^{2}+p^{2 }< be sure to add p^{2} to both sides
y^{2}2py+p^{2}=16p^{2}: Factor to find the square root for both sides
(yp)^{2 }= ±4p^{2}:
yp= +4p and yp = 4p
So. y = 5p and y = 3p
2. x^{2}3kx+2k^{2} = 0
x^{2}3kx+ = 2k^{2} : Divide the number in the middle term (3) by 2. That is (3/2) and square it...9/4 k^{2}
x^{2}3kx+9/4 k^{2}= 1/4 k^{2 }< Add 9/4 k^{2} to both sides
(x3/2 k)^{2 }= ±1/2 k
x3/2 k = 1/2 k and x3/2 k = 1/2 k
So, x = 2k and x = k
4/1/2014

Greg O.