Chapter 9: Problem 20
Solve by factoring and then solve using the quadratic formula. Check answers. $$y_{2}-1=0$$
Short Answer
Expert verified
The solutions are \( y = 1 \) and \( y = -1 \).
Step by step solution
01
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 \( c = -1 \).
02
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 \( y + 1 = 0 \).
03
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 \( y = -1 \).
04
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 \( y = -1 \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
Once factored, solve each expression by setting them to zero. This leads to two simple linear equations:
For the difference of squares, \( a^2 - b^2 \), the formula is:
- \( (a - b)(a + b) \)
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 \( y = -1 \)
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:
\[y = \frac{0 \pm \sqrt{0^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{\pm \sqrt{4}}{2} = \frac{\pm 2}{2}\]
This simplifies to \( y = 1 \) and \( y = -1 \).
The quadratic formula offers a reliable solution method by providing exact values for any quadratic equation.
\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula works by calculating the roots based on the coefficients \( a \), \( b \), and \( c \). For \( y^2 - 1 = 0 \), the values are:
- \( a = 1 \)
- \( b = 0 \)
- \( c = -1 \)
\[y = \frac{0 \pm \sqrt{0^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{\pm \sqrt{4}}{2} = \frac{\pm 2}{2}\]
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:
Understanding difference of squares not only makes factoring easier but also enhances problem-solving speed and accuracy.
- \( (a - b)(a + b) \)
- \( y^2 \) is \( a^2 \) and \( 1 \) is \( b^2 \)
Understanding difference of squares not only makes factoring easier but also enhances problem-solving speed and accuracy.