Question

find the value of p for which the roots of the equation px(x-2)+6=0are equal.

Answer 1

b^2-4ac=0 (== equa roots)
px(x-2)+6=0
px^2-2px+6=0
b= (-2p)
c=6
a=p
4p^2-24p=0
4p(p-6)=0
Then p is 0,6

Answer 2

hello users ...

Given that;
Roots of equation  
px(x-2)+6=0  Or px² - 2px + 6 = 0  Are equal.

We have to find p = ? 

Solution:-
we know that 
for a quadratic equation ax² + bx + c = 0
Roots are equal when 
 b² - 4ac = 0 

Here,
For equation px² - 2px + 6 = 0
by comparing ...
a = p , b = -2p and c = 6 

For equal roots 
b² - 4ac = 0 

=> (-2p)² - 4 * p * 6 = 0
=> 4p² - 24p = 0 
=> p² - 6p = 0 
=> p(p-6) = 0 
=> p = 0  and p - 6 = 0 
=> p = 0 and 6 Answer 

# hope it helps :)