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 =...

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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**