\[kx^2-17x+2k=0\]
Find the maximum integral value of k for which the above quadratic equation has two distinct real roots.
kx^2 - 17x + 2k = 0
If this has two real distinct roots, the discriminant must be > 0
So
(-17)^2 - 4(k)(2k) > 0
289 - 8k^2 > 0
289 > 8k^2
289/8 > k^2 take the positive root
sqrt (289/8) > k
≈ 6.01 > k
So....the max integral value for k = 6
