Question

find the value of p for which the given quadratic equation has equal roots: x²+p(2x-p-1)+2=0​

Answer 1

Answer:

I don't know

Step-by-step explanation:

Sorry I don't know answer

Answer 2

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