X2+(k+1)x+k =0 has root a and b find the value of k if a=2b

For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of roots is $-\frac{B}{A}$ and the product of roots is $\frac{C}{A}$.
Given the equation $x^2 + (k+1)x + k = 0$, we have $A=1$, $B=(k+1)$, and $C=k$.
The roots are $a$ and $b$.
Sum of roots:

$$a+b = -(k+1) \quad (*)$$

Product of roots:

$$ab = k \quad (**)$$

Substitute $a=2b$ into equation $(*)$:

$$2b+b = -(k+1)$$ $$3b = -(k+1) \quad (1)$$

Substitute $a=2b$ into equation $(**)$:

$$(2b)b = k$$ $$2b^2 = k \quad (2)$$

From equation (1), express $b$ in terms of $k$:

$$b = -\frac{k+1}{3}$$

Substitute this expression for $b$ into equation (2):

$$2\left(-\frac{k+1}{3}\right)^2 = k$$ $$2\frac{(k+1)^2}{9} = k$$ $$2(k^2 + 2k + 1) = 9k$$ $$2k^2 + 4k + 2 = 9k$$

$$2k^2 + 4k - 9k + 2 = 0$$ $$2k^2 - 5k + 2 = 0$$

Factor the quadratic equation:

$$2k^2 - 4k - k + 2 = 0$$ $$2k(k-2) - 1(k-2) = 0$$ $$(2k-1)(k-2) = 0$$

This gives two possible values for $k$:

$$2k-1 = 0 \implies 2k = 1 \implies k = \frac{1}{2}$$ $$k-2 = 0 \implies k = 2$$

The values of $k$ are $\frac{1}{2}$ and 2.

The value of $k$ is $\boxed{\frac{1}{2} \text{ or } 2}$.