Question

find the value of p for which the quaratic equestion x(x-4)+p=0has real root

Answer 1

Answer:

p≤4

Step-by-step explanation:

x(x-4)+p=0

x^2-4x+p=0

from the equation

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

for real roots

D≥0. (D=b^2-4ac)

(-4)^2-4(1)(p) ≥0

16-4p ≥0

16≥4p

4p≤16

p≤4

HOPE TO UNDERSTAND

PLEASE MARK AS BRILLIANT