determine the values of k so that 1 root is double the other root x^2-kx+7

x^2-kx +7= 0.

One root is 2
times the other.

To solve for x, we assume x1 and 2x1 are
the roots of the equation.

We know by the theory of
equations that if x1 and x2 are the roots of the equation f(x) = 0, then  f(x) =
k(x-x1)(x-x2).

Therefore x^2-kx+7 = k(x-x1)(x-2x1), where
we choose k =1 so that the coefficients of leading terms x^2 on both sides are
equal.

 x^2-kx+7
=(x-x1)(x-2x1)

x^2-kx+7 =
x^2-(x1+2x1)x+2x1^2

x^2-kx =
x^2-3x1+2x1^2

Equating x's and constanterms on both sides
we get:

-k = - 3x1

k =
3x1..........(1)

7 = 2x1^2.

So
x1^2= 7/2.

x1 = sqrt(7/2)   or
-sqrt(7/2).

So from (1) k = 3x1 = sqrt(7/2) . Or k =
-sqrt(7/2).