I need help with these two problems:

1. y

^{2}- 2py - 15p^{2}= 02. x

^{2}- 3kx + 2k^{2}= 0-
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I need help with these two problems:

1. y^{2} - 2py - 15p^{2} = 0

2. x^{2} - 3kx + 2k^{2} = 0

Tutors, please sign in to answer this question.

1. y^{2}-2py-15p^{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.

y2-2py+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

(y-p)^{2 }= ±4p^{2}:

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

(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

1. y^{2} - 2py - 15p^{2} = 0

(y-p)^{2}=y^{2}-2py+p^{2}

0=y^{2} - 2py - 15p^{2} =(y-p)^{2}-16p^{2}=(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. x^{2} - 3kx + 2k^{2} = 0

(x-3k/2)^{2}=x^{2}-3kx+9k^{2}/4

so that 0=x^{2} - 3kx + 2k^{2} =(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

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