x^2-kx+10=0

Solve for k:
10-k x+x^2 = 0

Multiply both sides by -1:
-10+k x-x^2 = 0

Add x^2+10 to both sides:
k x = x^2+10

Divide both sides by x:
Answer: | 
| k = x+10/x

x^2-kx+10=0

k can equal anything ://

if you want x to have real solutions then the determinant b^2-4ac must be greater than or equal to zero

\(k^2-40\ge0\\ k^2\ge40\\ k\ge\sqrt{40}\;\;\;or\;\;k\le-\sqrt{40}\\ approximate\; end\; points\\ k\ge6.32\;\;\;or\;\;k\le-6.32\\\)

If x is real and k is rational then k can be any number that makes the above inequality true.

Here is a graph.

https://www.desmos.com/calculator/kp8hozb4kl

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I suspect your question is poorly worded and the meaning is different but I am not sure what that would be ://

Maybe x and k BOTH have to be rational numbers.  Mmmm

WEll in this case there are still many answers but the determinant (b^2-4ac) must be a perfect square.

if k = 7   then  

x^2-7x+10=0

(x-2)(x-5)=0

x=2 or x=5

This is just one of many possible answers.

I

I chose 7 because k^2-40 must be a square number.

so k^2