Question

p^2+4y^2 if p-2y=2 and py=8​

Answer 1

Answer

Given equation:-

4x+py+8=0

a

1

=4,b

1

=p,c

1

=8

2x+2y+2=0

a

2

=2,b

2

=2,c

2

=2

Since the given pair of equation hs unique solution.

Therefore,

a

2

a

1



=

b

2

b

1

2

4



=

2

p

⇒p



=4

Hence the given pair of equation has unique solution for all values of p except 4.