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# Solving Equations by completing the square.

I need help with these two problems:

1. y2 - 2py - 15p2 = 0

2. x2 - 3kx + 2k2 = 0

### 2 Answers by Expert Tutors

Greg O. | Greg - Math and VolleyballGreg - Math and Volleyball
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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)= ±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 k<-- Add 9/4 k2 to both sides
(x-3/2 k)= ±1/2 k
x-3/2 k = 1/2 k and x-3/2 k = -1/2 k
So, x = 2k and x = k

Michael F. | Mathematics TutorMathematics Tutor
4.7 4.7 (6 lesson ratings) (6)
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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