P(x-4)(x-2) +(x-1)²=0

Given quadratic equation is p(x-4)(x-2)+(x-1)² = 0

=> p(x²-4x-2x+8)+(x²-2x+1) = 0

=> p(x²-6x+8)+(x²-2x+1) = 0

=> px²-6px+8p+x²-2x+1 = 0

=> (px²+x²)+(-6px-2x)+8p+1 = 0

=> (p+1)x²-(6p+2)x+(8p+1) = 0

On comparing with the standard quadratic equation is ax²+bx+c = 0 then

a = p+1

b = -(6p+2)

c = 8p+1

Given that

Given equation has real and equal roots

=> The discriminant = 0

We know that

The discriminant of ax²+bx+c = 0 is b²-4ac

Therefore, b²-4ac = 0

=> [-(6p+2)]²-4(p+1)(8p+1) = 0

=> (6p+2)²-4(p+1)(8p+1) = 0

=> (36p²+24p+4)-4(8p²+p+8p+1) = 0

=> (36p²+24p+4)-4(8p²+9p+1) = 0

=> 36p²+24p+4-32p²-36p-4 = 0

=> (36p²-32p²)+(24p-36p)+(4-4) = 0

=> 4p² -12p+0 = 0

=> 4p² - 12 p = 0

=> 4p(p-3) = 0

=> 4p = 0 or p-3 = 0

=> p = 0/4 or p = 3

=> p = 0 or p = 3

If p = 0 the term containing p doesn't exist .

Therefore, p = 3