What is the largest value of k such that the equation 6x - x^2 = k has at least one real solution?

Guest Oct 21, 2022

#1**+1 **

x^2-6x+k=0

\(x = {-(-6) \pm \sqrt{(-6)^2-4(1)(k)} \over 2(1)}=3\pm1\sqrt{9-k}\)

If k gets any bigger than 9, there both roots will be zero. Hence the largest value k can be such that that equation has at least one real solution is 9.

WhyamIdoingthis Oct 21, 2022