find the value of p for which the given quadratic equation has equal roots: x²+p(2x-p-1)+2=0 About the author Skylar
x²+p(2x-p+1)+2=0 x²+2px-p²+p+2=0 x²+2px-1(p²-p-2)=0 where a=1 b= 2p c= -(p²-p-2) if this is a quadratic equation with equal roots then b²-4ac= 0 (2p)²- 4(1×-(p²-p-2)= 0 4p²-4(-p²+p+2) 4p²+4p²-4p-8= 0 8p²-4p-8= 0 4(2p²-p-2)=0 2p²-p-2=0 Here, again a= 2 b=- 1 c=-2 So D = d²= b²-4ac= 1-4(2×-2)= 1-(-16)= 1+16= 17 now p= -b//2a ± d/2a where d= √17 = –(–1) ± √17 2(2) 2(2) = 1 ± √17 4 4 = 1 ± √17 4 Reply
Answer:
I don’t know
Step-by-step explanation:
Sorry I don’t know answer
x²+p(2x-p+1)+2=0
x²+2px-p²+p+2=0
x²+2px-1(p²-p-2)=0
where a=1
b= 2p
c= -(p²-p-2)
if this is a quadratic equation with equal roots then
b²-4ac= 0
(2p)²- 4(1×-(p²-p-2)= 0
4p²-4(-p²+p+2)
4p²+4p²-4p-8= 0
8p²-4p-8= 0
4(2p²-p-2)=0
2p²-p-2=0
Here, again a= 2
b=- 1
c=-2
So D = d²= b²-4ac= 1-4(2×-2)= 1-(-16)= 1+16= 17
now p= -b//2a ± d/2a where d= √17
= –(–1) ± √17
2(2) 2(2)
= 1 ± √17
4 4
= 1 ± √17
4