The value of $ \gamma $ in order that the equations $ 2x^{2}+5\gamma x+2=0 $ and $ 4x^{2}+8\gamma x+3=0 $ have a common root is given by

Let $\alpha$ be the common root of the equations $2x^2 + 5\gamma x + 2 = 0$ and $4x^2 + 8\gamma x + 3 = 0$. Multiplying the first equation by $2$ gives $4\alpha^2 + 10\gamma \alpha + 4 = 0$, and subtracting the second equation yields $2\gamma \alpha = 1$, so $\alpha = 1 / (2\gamma)$. Substituting this back into the first original equation gives $2(1 / 4\gamma^2) + 5\gamma(1 / 2\gamma) + 2 = 0$, which simplifies to $1 / 2\gamma^2 + 9/2 = 0$. Solving $1 + 9\gamma^2 = 0$ results in $\gamma^2 = \pm 1$, so $\gamma = \pm 1$.

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