Question

. The value of p for which the roots of the quadratic equation x2-(p-2)x+p+1=0 are real and equal is

Answer 1

Answer:

P = 8

Step-by-step explanation:

Nature of roots are real and equal if D = 0

where  $D = b^{2} - 4ac$

given equation,$x^{2} - (p-2)x+p+1=0$ is of the form $ax^{2} + bx+c=0$

here; a = 1, b = - (p - 2), c = p+1

$D = b^{2} - 4ac =0 \\\\ D = [-(P-2)]^{2} - 4(1)(P+1) = 0\\\\ P^{2} +4-4P-4(P+1)=0\\\\ P^{2} +4-4P-4P-4=0\\\\ P^{2} -8P=0\\\\P^{2}=8P\\\\ P=8$