Question

For how many positive integers for k is the value of k^2 - 18k + 65 a positive prime number?
0, 1, 2, or 4?

Answer 1

see if you can factorise the expression:\[k^2-18k+65\]if you can then this implies it can only generate a prime number if one of the factors is equal to 1.

Answer 2

it factors into
(x-5)(x-13)
and how do you know that it can only generate a prime number if its 1? Oh! Any integer squared is automatically composite unless its 1. So we try 1 and -1 and we get 48 and 84 respectively. So, the answer is 0? But even if something is squared and it is composite, it could still be changed into prime by the -18 or the + 65

Answer 3

Well, not by the -18, but it could indirectly help the 65 make f(k) prime.

Answer 4

In general x*y is composite (by definition) unless x=1 and y is prime or y=1 and x is prime.
for your case (with integer k>0)
(k-5)(k-13)
we ask can (k-5) = 1 and (k-13) be prime and vice versa
for k=4, we get -1* -9 = 9 not prime
k=6 1* -7 negative but the question asks for positive primes
k= 12 7 * -1 neg
k=14 9 * 1 not prime

Answer 5

It looks like zero primes

Answer 6

How come we want to make the x-intercepts = 1? Doesnt that leave out a whole bunch of other numbers that we could try into the original equation?

Answer 7

the answer is 0 but i dont understand how you arrived at that

Answer 8

Do you agree with this statement:
x*y is composite (by definition) unless x=1 and y is prime or y=1 and x is prime.

Answer 9

k^2 - 18 k + 65 = (k-5)(k-13)

Answer 10

Yes, because after the multiplication, that product would be divisible by both x and y

Answer 11

so x=(k-5) and y= (k-13)

Answer 12

?

Answer 13

x*y is composite unless....
(k-5)*(k-13) is composite unless....

Answer 14

Unless one of those equals 1! Oh. I see now. Goodness I can be a real stonehead sometimes heheh. Thanks

Answer 15

I was getting nervous, cause I do not know how to say it any other way