Problem 84 - Algebra for College Students

Question

Find all real or imaginary solutions to each equation. Use the method of your choice. $$\left(y-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Step-by-Step Solution

Answer

The solutions are \(y = \frac{4}{3} \) and \(y = 0 \).

Step 1: Understand the Equation

The given equation is \((y - \frac{2}{3})^2 = \frac{4}{9}\). This is a quadratic equation in the form \((a - b)^2 = c\). Our goal is to solve for \y\.

Step 2: Take the Square Root of Both Sides

To eliminate the square, apply the square root to both sides of the equation: \(\sqrt{(y - \frac{2}{3})^2} = \sqrt{\frac{4}{9}}\). This simplifies to \(|y - \frac{2}{3}| = \frac{2}{3}\).

Step 3: Set Up Two Separate Equations

Since \(|y - \frac{2}{3}| = \frac{2}{3}\) implies two possible scenarios, we can write two equations: \( y - \frac{2}{3} = \frac{2}{3}\) and \( y - \frac{2}{3} = -\frac{2}{3}\).

Step 4: Solve Each Equation

Solve the first equation: \( y - \frac{2}{3} = \frac{2}{3} \Rightarrow y = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \). Solve the second equation: \( y - \frac{2}{3} = -\frac{2}{3} \Rightarrow y = \frac{2}{3} - \frac{2}{3} = 0 \).

Step 5: Write the Final Solutions

Thus, the solutions to the equation \((y - \frac{2}{3})^2 = \frac{4}{9}\) are \(y = \frac{4}{3} \) and \(y = 0 \).

Key Concepts

Imaginary Solutions

In algebra, some quadratic equations may not have real solutions. This happens when the solutions involve the square roots of negative numbers. These are called imaginary solutions because they do not correspond to any point on the real number line.
Imaginary numbers are built around the imaginary unit, represented by \text{\(i\)}, where \text{\(i^2 = -1\)}. For example, the square root of \text{\( - 4\)} is \text{\(2i\)}.
When solving quadratic equations, if you find yourself needing to take the square root of a negative number, this is where you will get imaginary solutions. You can use the quadratic formula or complete the square method to find these solutions.
Remember, imaginary solutions always come in pairs: \text{\(a + bi\)} and \text{\(a - bi\)}.

Simplifying Equations

Simplifying equations is crucial to make them easier to solve. Simplification can involve combining like terms, canceling out terms, or dividing through by a common factor.
In the given exercise, we started with \text{\((y - \frac{2}{3})^2 = \frac{4}{9}\)}. By taking the square root of both sides, we simplified it to \text{\(|y - \frac{2}{3}| = \frac{2}{3} \)}. This step removes the square, making the equation easier to handle.
Simplifying equations can also involve making them more visually understandable. For example, converting fractions like \text{\( \frac{4}{9}\)} to their simplest form or using factoring techniques.

Square Root Method

The square root method is a technique used to solve quadratic equations. When you have an equation in the form \text{\((a - b)^2 = c\)}, you can use this method. Here’s how:

• First, take the square root of both sides to remove the square. This gives you \text{\(|a - b| = \sqrt{c}\)}.
• Next, set up two equations because the absolute value means there are two possible cases: \text{\(a - b = \sqrt{c}\)} and \text{\(a - b = -\sqrt{c}\)}

For our example, taking the square root of \text{\((y - \frac{2}{3})^2 = \frac{4}{9}\)} leads us to \text{\(|y - \frac{2}{3}| = \frac{2}{3}\)}.
Then, solving each of these separate equations, we get the solutions: \text{\(y = \frac{4}{3}\)} and \text{\(y = 0\)}.

Absolute Value Equations

Absolute value equations involve expressions within an absolute value symbol \text{\(| \cdot |\)}. The absolute value of a number is its distance from zero on the number line, disregarding the direction.
For an equation in the form \text{\(|A| = B\)}, set up two equations to account for both scenarios: A could equal B, or A could equal -B.

• From \text{\(|y - \frac{2}{3}| = \frac{2}{3}\)}, we get two cases: \text{\(y - \frac{2}{3} = \frac{2}{3}\)} and \text{\( y - \frac{2}{3} = - \frac{2}{3}\)}
• Solve each case separately to find the complete set of solutions.

Using these methods helps break down more complicated problems into manageable steps, leading to a final solution.