Which of the following four equations has the solution of the lowest value?
To determine which equation has the lowest solution, we need to find the lowest value for the solution. We can solve each equation and compare the solutions:
1) $x^2 - 3x - 4 = 0$
To solve this quadratic equation, we can factor it as $(x-4)(x+1) = 0$. This gives us two solutions: $x=4$ and $x=-1$. The lowest value for $x$ is $-1$.
2) $2x^2 + 3x + 1 = 0$
We can solve this quadratic equation by factoring as $(2x+1)(x+1) = 0$. This gives us two solutions: $x=-1$ and $x=-\frac{1}{2}$. The lowest value for $x$ is $-1$.
3) $3x^2 + 5x + 2 = 0$
We can solve this quadratic equation by factoring as $(3x+1)(x+2) = 0$. This gives us two solutions: $x=-\frac{1}{3}$ and $x=-2$. The lowest value for $x$ is $-2$.
4) $4x^2 - 2x - 1 = 0$
We can solve this quadratic equation using the quadratic formula, which gives us $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-1)}}{2(4)}$. Simplifying this gives us $x = \frac{1 \pm \sqrt{10}}{4}$. The lowest value for $x$ is approximately $x \approx -0.68$.
Comparing the lowest values for $x$ in each equation, we find that the equation with the lowest solution is $\boxed{1) \ x^2 - 3x - 4 = 0}$.