\(f(x)= x^2 - (k+8)x + (8k+1)\)

Show that when k =8, f(x)>0 for all values of x

YEEEEEET Oct 20, 2018

#1**+2 **

When k = 8 we have \(f(x) = x^2-16x+64+1 \text{ or } f(x) = (x - 8)^2 + 1\)

No matter what the value of x is, (x-8)^{2} is always zero or positive, Hence f(x) is always greater than zero.

Alan Oct 20, 2018