Solving Equations by completing the square.

Question

I need help with these two problems:

 

1. y² - 2py - 15p² = 0

 

2. x² - 3kx + 2k² = 0

Answer 1

1. y²-2py-15p²=0

y²-2py+      =15p²: Then, Divide the coefficient of y in the middle term (-2p) by 2 (that will be -1p) and square it. You get 1p² or p² (they are the same). That is completing the square.

p² is your new third term.

y2-2py+p²=15p²+p²  <-- be sure to add p² to both sides

y²-2py+p²=16p²: Factor to find the square root for both sides

(y-p)² = ±4p²:

y-p= +4p and y-p = -4p

So. y = 5p and y = -3p

 

2. x²-3kx+2k² = 0

x²-3kx+     = -2k² : Divide the number in the middle term (-3) by 2. That is (-3/2) and square it...9/4 k²

x²-3kx+9/4 k²= 1/4 k² <-- Add 9/4 k² to both sides

(x-3/2 k)² = ±1/2 k

x-3/2 k = 1/2 k and x-3/2 k = -1/2 k

So, x = 2k and x = k

 

 

Answer 2

1. y² - 2py - 15p² = 0

(y-p)²=y²-2py+p²

0=y² - 2py - 15p² =(y-p)²-16p²=(y-p)²-(4p)²=(y-p+4p)(y-p-4p)=(y+3p)(y-5p)

For which we have that y=-3p or that y=5p

 

2. x² - 3kx + 2k² = 0

(x-3k/2)²=x²-3kx+9k²/4

so that 0=x² - 3kx + 2k² =(x-3k/2)²-((1/2)k)²=(x-3k/2-(1/2)k)(x-3k/2+(1/2)k)

0=(x-2k)(x-k) so that x=2k or x=k

 

This is a good exercise for learning the quadratic formula