Question

find the value of k for which equation x square +kx +10=0 has real roots

Answer 1

Answer:

Step-by-step explanation:

x²-kx+10=0

Decrimination=b²-4ac

=(-k)²-4(1)(10)

=k²-40

for roots to be real

b²-4ac>0

or

k2-40>0

K2>40

k>±2√10

k>2√10 and k<-2√10