91Ó°ÊÓ

In solving equations, isolating the variable is a key step. This means you get the variable (like t) by itself on one side of the equation. Let's see this in action with the equation \(\bigg(t - \frac{5}{6}\bigg)^2 = \frac{11}{25}\). Here, our variable t is inside a square. We want to get t alone, so we'll start by taking the square root of both sides. This isolates the expression with our variable. We get: \(\big(t - \frac{5}{6}\big) = \pm \sqrt{\frac{11}{25}}\). Remember, whenever you take the square root of both sides of an equation, you need to consider both the positive and negative roots. Now the variable t is almost isolated; we just need to handle the \(- \frac{5}{6}\) next.

Using the square root property can seem tricky at first but it is a helpful tool. The square root property tells us if \(a^2 = b\), then \(a = + \sqrt{b}\) or \(a = - \sqrt{b}\). Applying this to our equation: \(\big(t - \frac{5}{6}\big)^2 = \frac{11}{25}\), we take the square root of both sides. This gives us two possible equations: \(t - \frac{5}{6} = \frac{\sqrt{11}}{5}\) and \(\big(t - \frac{5}{6}\big) = - \frac{\sqrt{11}}{5}\). Now we see that our variable t is in two places with different values. These are the possible solutions we need to explore.

Rationalizing fractions might sound complex, but it's just a way to make sure our answers are in a simple form without any roots in the denominator. In our equation \(t = \frac{5}{6} \pm \frac{\sqrt{11}}{5}\), everything looks pretty clean already. But if we had \frac{1}{\frac{\sqrt{2}}}\ in a problem, we'd want to rationalize it by multiplying the numerator and the denominator by the square root to get \frac{\textrm{\tiny{\sqrt{2}}}}{2}\. Rationalizing makes our equations easier to understand and any further calculations simpler. Our final answer after isolating the variable and using the square root property leaves us with neat, rational solutions.