Question

Find p for equation px (x-3) + 9=0 have real and equal roots

Answer 1

Answer:

PX²-3PX +9=0

hope its Right yar......

Answer 2

Answer:

Given that Equation has real and equal roots means D=b^2 -4ac =0

now, px(x-3)+9=0

px^2 -3px +9=0, so we have, a=p, b=-3p and c=9

(-3p)^2 -4*p*9=0

9p^2 -4p*9=0

p^2 -4p=0

p(p-4)=0

p=0 and p-4=0

p=0 and p=4