Question

in the following determine the set of the value of p for which give quadratic equation has real roots.1) x^2-4x+p=0 (2) 9x^2-5x+(p+1)=0

Answer 1

D = b2 - 4ac

a = 1 , b = -4 , c = p

16 - 4p = 0

-4p = -16

p = 4

a = 9 , b = -5, c = p+1

25 - 4(9)(p+1) = 0

25 - 36p + 36 = 0

-36 p = -36-25

p = -61/-36

p= = 61/36