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 $\boxed{1) \ x^2 - 3x - 4 = 0}$.
Which of the following four equations has the solution of the lowest value?
1 answer