Question

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

Answer 1

Given quadratic equation is= px (x-3)+9.
px² -3px +9= 0

On comparing with standard form of quadratic equation i.e ax² + bx + c =0, a≠ 0

Here, a = p , b= -3p, c= 9
D(discriminant)= b²-4ac
= (-3p)² - 4× p× 9
= 9p² - 36p
= 9p(p -4)

Since, roots of given equation are real and equal. D= 0

0 = 9p(p -4)
9p= 0 or  p - 4= 0
p ≠ 0 or p = 4
Hence, required value of p is 4

HOPE THIS WILL HELP YOU...

Answer 2

therefore p=4........