Question

find the value of p for which the quadratic equation px(x-3)+9=0has equal roots????

Answer 1

 px(x-3) + 9 = 0

px2-3px+9=0

a=p , b=-3p , c=9

b2-4ac= (-3p)2-4(p)(9)

=9p2-36p

for having equal roots 

b2-4ac = 9p2-36p = 0

9p2-36p = 0

9p(p-4) = 0

9p=0 (or) p-4=0

p=0(rejected) or p=4

therfour, the value of p=4

hope u got it ....:D

Answer 2

$p {x}^{2} - 3px + 9 = 0$
sum of roots for the quadratic equation = -b/a
$a {x}^{2} + bx + c = 0$
So sum of roots of given equation
-(-3p/p)= 3

roots are equal so roots will be 3/2, 3/2

substitute this in given equation, then we will get....

$p = 4$