Question

Find the value of p for which the equation has equal roots px(x-2) +6

Answer 1

px(x+2)+6

px^2+2px+6

here a=p, b=2p ,c=6

b^2-4ac=(2p)^2-4(2p)(6)

=4p^2-48p

As they have real and equal roots the value of ∆=b^2-4ac=0

∆=0

∆=4p^2-48p

0=4p^2-48p

0=4p(p-48)

0=4p or p-48=0

p=0/4 or p=+48

p=0 or p=48

The value of p can't be 0 if it is 0 then it will be not a quadratic eqn

Therefore value of p is 48