Problem 20 - Elementary Algebra

Question

Solve by factoring and then solve using the quadratic formula. Check answers. $$y_{2}-1=0$$

Step-by-Step Solution

Answer

The solutions are \( y = 1 \) and \( y = -1 \).

Step 1: Rewrite the Equation

Start by rewriting the equation in standard form. The given equation is \( y^2 - 1 = 0 \). This is already in the standard quadratic form \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 0 \), and \( c = -1 \).

Step 2: Solve by Factoring

Recognize that \( y^2 - 1 \) is a difference of squares, which can be factored as \( (y - 1)(y + 1) = 0 \). Set each factor equal to zero: \( y - 1 = 0 \) and \( y + 1 = 0 \).

Step 3: Solve Factored Equations

Solve each individual equation for \( y \):- For \( y - 1 = 0 \), add 1 to both sides to get \( y = 1 \).- For \( y + 1 = 0 \), subtract 1 from both sides to get \( y = -1 \).

Step 4: Solve Using the Quadratic Formula

Recall the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = 0 \), and \( c = -1 \) into the formula: \[y = \frac{0 \pm \sqrt{0^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{\pm \sqrt{4}}{2} = \frac{\pm 2}{2} \]This simplifies to \( y = 1 \) and \( y = -1 \).

Step 5: Check Answers

Substitute \( y = 1 \) and \( y = -1 \) back into the original equation to verify the solutions:- For \( y = 1 \), \( y^2 - 1 = 1^2 - 1 = 0 \), which is correct.- For \( y = -1 \), \( y^2 - 1 = (-1)^2 - 1 = 0 \), which is also correct.

Key Concepts

Factoring

Factoring is a powerful method used to solve quadratic equations by transforming them into a product of simpler expressions. When we factor a quadratic like \( y^2 - 1 = 0 \), we essentially rewrite it as a product of two binomials. Here, the equation \( y^2 - 1 \) is in the form of a difference of squares, which allows us to factor it using a special formula.

For the difference of squares, \( a^2 - b^2 \), the formula is:

• \( (a - b)(a + b) \)

This means \( y^2 - 1 \) can be factored to \( (y - 1)(y + 1) = 0 \).

Once factored, solve each expression by setting them to zero. This leads to two simple linear equations:

• \( y - 1 = 0 \) giving \( y = 1 \)
• \( y + 1 = 0 \) giving \( y = -1 \)

Thus, factoring helps us break down complex quadratics into straightforward solutions.

Quadratic Formula

The quadratic formula is a universal method to find solutions for any quadratic equation. It's especially useful when factoring is difficult or impossible. The standard form of a quadratic is \( ax^2 + bx + c = 0 \) and the solutions are given by:

\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula works by calculating the roots based on the coefficients \( a \), \( b \), and \( c \). For \( y^2 - 1 = 0 \), the values are:

• \( a = 1 \)
• \( b = 0 \)
• \( c = -1 \)

Plug these into the formula:

\[y = \frac{0 \pm \sqrt{0^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{\pm \sqrt{4}}{2} = \frac{\pm 2}{2}\]
This simplifies to \( y = 1 \) and \( y = -1 \).

The quadratic formula offers a reliable solution method by providing exact values for any quadratic equation.

Difference of Squares

The difference of squares is a specific method used in factoring that applies only to expressions in a particular form. In algebra, a difference of squares is described as \( a^2 - b^2 \). This expression can always be factored into:

• \( (a - b)(a + b) \)

Recognizing this pattern is crucial for quickly solving equations like \( y^2 - 1 \), where:

• \( y^2 \) is \( a^2 \) and \( 1 \) is \( b^2 \)

Thus, \( y^2 - 1 = (y - 1)(y + 1) \). This method streamlines solving for \( y \) since each factor leads immediately to a simple solution. When encountering a difference of squares, it is a straightforward path to quickly factor and find the solutions to the quadratic equation.

Understanding difference of squares not only makes factoring easier but also enhances problem-solving speed and accuracy.