Question

Find value of p so that quadratic equation px(x-3)+9=0 has tow equal roots

Answer 1

b^2-4ca=0;............................(1)
px^2-3pc+9=0;
a=p;
b=-3p
c=9
substitute:
9p^2-36p=0;
9p^2=36;
p^2=3p;hence:
p=3

Answer 2

HERE IS YOUR ANSWER MATE.....;

$= > b {}^{2} - 4ac = 0$

$= > ( - 3p) {}^{2} - 4 \times p \times 9 = 0$

$= > 9p {}^{2} - 36p = 0$

$= > p {}^{2} - 4p = 0$

$= > p(p - 4) = 0$

$= > p - 4 = \frac{0}{p} \: \: \:(here \:, \frac{0}{p} = 0)$

$= > p = + 4$

HOPE IT'S HELPFUL....:-)