Determine the value of k for which the equation 4x^2+kx+4=0 has one root equal to double the other root.

Let the value of one of the roots be R, the other root is
2R.

Now the quadratic equation can be derived by the
following operation: (x - R)(x - 2R) = 0

=> x^2 - xR
- 2xR + 2R^2 = 0

=> x^2 - 3xR + 2R^2 =
0

This is equivalent to 4x^2 + kx + 4 =
0

Dividing this by 4

=>
x^2 + (k/4)x + 1 = 0

Now equate the coefficients of the x^2
- 3xR + 2R^2 = 0 and x^2 + (k/4)x + 1 = 0

We get k/4 = -3R
and 2R^2 = 1

2R^2 =
1

=> R^2 =
1/2

=> R = 1/ sqrt 2 or -1/ sqrt
2

substitute this in k/4 =
-3R

=> k = -12*(1/sqrt 2) or k = 12/sqrt
2

=> k = 6 sqrt 2 or k = -6 sqrt
2

Therefore k can take the values k = 6 sqrt
2 and k= -6 sqrt 2