91Ó°ÊÓ

A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a eq 0\). In our exercise, the function given is \(F(t) = (t + 4)^2\). Quadratic equations are fundamental in algebra and can be used to describe parabolic patterns seen in various real-world contexts. The standard method to solve such equations often involves isolating the variable.
To find the solutions for our specific equation \((t + 4)^2 = 13\), we treated it as a quadratic equation. The challenge here was to work with the expression inside the parentheses and solve for the variable 't'.
Always remember, a key step in dealing with quadratic equations is to set the equation to 0 if not already done, before solving for the variable.

Solving equations involves finding the value of the unknown variable that makes the equation true. In this exercise, the goal was to find \(t\) such that \(F(t) = 13\). We started by setting up the equation \((t + 4)^2 = 13\). This initial step is crucial as it translates the problem statement into a mathematical expression we can work with.
The next step was to take the square root of both sides, which helps to get rid of the square on the left-hand side. Here, we used \(\sqrt{(t + 4)^2} = \pm \sqrt{13}\). The symbol \(\pm\) indicates that there are two possible values for \(\sqrt{13}\). This simplification gave us the equation \(t + 4 = \pm \sqrt{13}\).
The final step was to solve for \(t\) by isolating it. We did this by subtracting 4 from both sides, leading to \(t = -4 \pm \sqrt{13}\). Thus, we had two solutions: \(t = -4 + \sqrt{13}\) and \(t = -4 - \sqrt{13}\). Understanding these sequential steps is key to mastering solving equations.

A square root is a value that, when multiplied by itself, gives the original number. The square root symbol \(\sqrt{}\) denotes this. In our exercise, taking the square root of both sides was a critical step. We had \(\sqrt{(t + 4)^2} = \pm \sqrt{13}\).
Here, \(\pm\) is essential because every positive number has two square roots: one positive and one negative. For example, the number 9 has two square roots, +3 and -3, because \(3^2 = 9\) and \((-3)^2 = 9\).
It's important to note that while working with quadratic equations, taking the square root of both sides always introduces these two possible values. This step is what gave us the two solutions for \(t\): \(t = -4 + \sqrt{13}\) and \(t = -4 - \sqrt{13}\). Always remember this when solving similar problems.