Problem 22 - Elementary Algebra

Question

Solve the following quadratic equations. \(\left(t-\frac{5}{6}\right)^{2}=\frac{11}{25}\)

Step-by-Step Solution

Answer

The solutions are \( t = \frac{5}{6} + \frac{\sqrt{11}}{5} \) and \( t = \frac{5}{6} - \frac{\sqrt{11}}{5} \).

Step 1: Understand the equation

The given equation is \((t - \frac{5}{6})^{2} = \frac{11}{25}\). This is a quadratic equation in the form \( (x - a)^2 = b \).

Step 2: Take the square root of both sides

To solve for \( t \), take the square root of both sides of the equation:\[(t - \frac{5}{6}) = \pm \sqrt{\frac{11}{25}}\].

Step 3: Simplify the square root

Simplify the square root on the right-hand side:\[\sqrt{\frac{11}{25}} = \frac{\sqrt{11}}{\sqrt{25}} = \frac{\sqrt{11}}{5}\].

Step 4: Set up two separate equations

We now have two equations to solve because of the \( \pm \) sign:\[(t - \frac{5}{6}) = \frac{\sqrt{11}}{5}\]and\[(t - \frac{5}{6}) = -\frac{\sqrt{11}}{5}\].

Step 5: Solve for t in both equations

Solve each equation separately:For \((t - \frac{5}{6}) = \frac{\sqrt{11}}{5}\)\[t = \frac{5}{6} + \frac{\sqrt{11}}{5}\].For \((t - \frac{5}{6}) = -\frac{\sqrt{11}}{5}\)\[t = \frac{5}{6} - \frac{\sqrt{11}}{5}\].

Key Concepts

quadratic equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. These equations form a parabola when plotted on a graph.
In our exercise, the equation given is \((t - \frac{5}{6})^{2} = \frac{11}{25}\), which can be seen as a special form of a quadratic equation. Solving quadratic equations can involve different methods: factoring, completing the square, using the quadratic formula, or, as in our problem, the square root method.

square root method

The square root method is a straightforward way to solve quadratic equations that are already in the form \((x - a)^2 = b\). Here’s how it works:

• First, identify the square on one side of the equation. In our problem, \((t - \frac{5}{6})^{2} = \frac{11}{25}\).
• Next, take the square root of both sides of the equation. Remember to include both the positive and negative roots. This becomes \(t - \frac{5}{6} = \pm \sqrt{\frac{11}{25}}\).

This method is efficient when the quadratic equation can be easily rewritten to match the square form. By taking the square root, we simplify our work significantly.

simplifying radicals

Simplifying radicals is often necessary when solving quadratic equations that involve square roots. In the given exercise, you take the square root of \(\frac{11}{25}\):

• First, note that \(\frac{11}{25}\) can be broken down under the square root: \(\sqrt{\frac{11}{25}} = \frac{\sqrt{11}}{\sqrt{25}}\).
• Simplify what you can inside the root. Since \( \sqrt{25} = 5 \), we get \(\frac{\sqrt{11}}{5}\).

Thus, the equation transforms to \(t - \frac{5}{6} = \pm \frac{\sqrt{11}}{5}\). Simplifying radicals makes the subsequent arithmetic and problem-solving steps more manageable.

solving for variables

Solving for variables involves isolating the variable on one side of the equation. This often requires performing inverse operations to undo any addition, subtraction, multiplication, or division affecting the variable.

• In our exercise, we have two separate equations due to the ± sign: \(t - \frac{5}{6} = \frac{\sqrt{11}}{5}\) and \(t - \frac{5}{6} = -\frac{\sqrt{11}}{5}\).
• For both equations, we add \(\frac{5}{6}\) to both sides to isolate \(t\).
• This gives us two solutions: \(t = \frac{5}{6} + \frac{\sqrt{11}}{5}\) and \(t = \frac{5}{6} - \frac{\sqrt{11}}{5}\).

By solving for \(t\) in both cases, we find the potential values that satisfy the original quadratic equation. Always ensure all necessary steps are followed to isolate and solve for the variable accurately.