Question

Tentukan nilai r pada persamaan bentuk aljabar (2x + 3y)(px + qy)= rx² + 23xy + 12y²

Answer 1

(2x+3y)(Px+Qy) = (rx^2+23xy+12y^2)
2Px^2+3Pxy+2Qxy+3Qy^2 = (rx^2+23xy+12y^2)
2Px^2+xy(3P+2Q)+3Qy^2 = (rx^2+23xy+12y^2)

Maka:
3Q = 12
Q = 4

3P+2Q = 23
3P + 2.4 = 23
3P = 23-8
3P = 15
P = 5

Daan...
2P = r
2.5 = r
r = 10

Answer 2

=2px^2+2qxy + 3pxy+3qy^2=rx^2+23xy+12y^2
=2px^2+(2q+3p)xy+3qy^2=rx^2+23+12y^2
=2px^2 =rx^2
=x^2 nya mati
=2p = r
=2q+p=23
=3q = 12

q=12/3
q=4
2q+3p=23
2.4+3p=23
3p=23+8
3p=31
p=31/3
p =10,3
r=10,3.2=20.6

NB : ^2 = PANGKAT 2