In mathematics, particularly in algebra and calculus, you might often encounter situations where you have to discard one of the solutions to an equation. This isn’t arbitrary but usually based on context and constraints related to the problem.
Contextual Relevance
Consider a quadratic equation like $x^2 – 5x + 6 = 0$. Solving this, we get two solutions: $x = 2$ and $x = 3$. But what if the problem states that $x$ must be greater than 2? In this case, we discard $x = 2$ because it doesn’t meet the condition.
Real-World Constraints
Sometimes, the nature of the problem imposes constraints. For instance, if you’re solving for the length of a side of a triangle, the solution must be a positive number. Suppose you solve an equation and get $x = 4$ and $x = -2$. Since a negative length doesn’t make sense, you discard $x = -2$
Example in Physics
In physics, when solving for time, you might get two solutions from a quadratic equation. One might be negative and the other positive. Since time can’t be negative in this context, you discard the negative solution.
Quadratic Equation Example
Let’s take the equation $t^2 – 4t – 5 = 0$. Solving it using the quadratic formula $t = frac{-b pm sqrt{b^2 – 4ac}}{2a}$, we get:
$t = frac{4 pm sqrt{16 + 20}}{2} = frac{4 pm sqrt{36}}{2} = frac{4 pm 6}{2}$
Thus, $t = 5$ and $t = -1$. Since time can’t be negative, we discard $t = -1$
Domain Restrictions
In trigonometry, you might solve an equation like $sin(x) = 0.5$. The general solutions are $x = frac{pi}{6} + 2kpi$ and $x = frac{5pi}{6} + 2kpi$ for any integer $k$. However, if the problem restricts $x$ to $[0, pi]$, only $x = frac{pi}{6}$ and $x = frac{5pi}{6}$ are valid.
Conclusion
Discarding one of the solutions is a crucial step in problem-solving. It’s based on the context, constraints, and domain restrictions of the problem. Understanding why a solution is discarded helps ensure that the answer makes sense and is applicable to the real-world scenario or mathematical context.
3. Physics Classroom – Solving Equations