What is the sum of all values of k such that the equation 2x^2-kx+8=0 has two distinct integer solutions?

k > 8

Step-by-step explanation:

Step 1: We know in order for a quadratic equation to have 2 distinct solutions the discriminant has to be positive

Important formula: Discriminant =

Step 2: Input information into discriminant

> 0

> 0

> 0

> 64

k > 8

Therefore in order for the equation to have 2 distinct solutions is to have k > 8

(b²-4ac) > 0

where Z is an integer

(-k)²-4(2)(8) > 0

k²-64 > 0

k²>64

k>8

Therefore the sum of all values of k is infinite

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb