Find the value of 'p' for which the quadratic equation p(x-4)(x-2) . . .

Question :  Find the value of 'p' for which the quadratic equation p(x-4)(x-2)+(x-1)²=0 has real and equal roots.

Doubt by CBSE

Solution : 

p(x-4)(x-2)+(x-1)²=0

p(x²-2x-4x+8)+(x)²+(1)²-2(x)(1)=0
p(x²-6x+8)+x²+1-2x=0
px²-6px+8p+x²+1-2x=0
px²+x²-6px-2x+8p+1=0
(p+1)x²-2(3p+1)x+(8p+1)=0

Now 
a = p+1
b = -2(3p+1)

c = 8p+1

D=b²-4ac

Roots are real and equal. 
So D=0

0=[-2(3p+1)]²-4(p+1)(8p+1)
0=4(3p+1)²-4(p+1)(8p+1)

0=4[(3p+1)²-(p+1)(8p+1)]
0/4=(3p+1)²-(p+1)(8p+1)

0=9p²+1+6p-[8p²+p+8p+1]

0=9p²+1+6p-[8p²+9p+1]
0=9p²+1+6p-8p²-9p-1
0=p²-3p
0=p(p-3)
p=0 or p-3=0
p=0 or p=3

Hence, p=0 OR 3