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:

Step-by-step explanation:

For real and equal roots, Determinant = b² – 4ac = 0.

So in the equation x²- (p-2)x + (p+1) = 0

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

=> [-(p - 2)] ² - 4 * 1 * (p+1) = 0

=> (p-2)² = 4(p+1)

=> p² - 4p + 4 = 4p + 4

=> p² =  8p

=> p = 8