Question #1cabf

1 Answer
Oct 8, 2017

a) #k^2-16k+60#
b) #k=6, 10#
c) as in explanation

Explanation:

a) in #ax^2+bx+c=0 (b^2-4ac)=>(-k-8)^2-4*1*(8k+1)=k^2-16k+60#
#b) b^2-4ac=0#=>two equal roots=>#k^2-16k+60=0=>(k-6)(k-10)=0=>#k=6 or k=10
c) when #b^2-4ac<0, f(x)# is always more than 0
so if #k=8, b^2-4ac=-4<0#, and #f(x)>0# always (when #a>0)#