Chapter 10: Problem 6
For the following problems, solve the equations. $$ (4 y+1)(2 y+3)=0 $$
Short Answer
Expert verified
Answer: The solutions for the equation are y = -1/4 and y = -3/2.
Step by step solution
01
Set each factor equal to zero
Since we have \((4y + 1)(2y + 3) = 0\), in order to find the solutions for y, we need to set each factor to zero and solve for y:
1. \((4y + 1) = 0\)
2. \((2y + 3) = 0\)
02
Solve the first equation
To solve the first equation \((4y + 1) = 0\), subtract 1 from both sides of the equation:
\(4y = -1\)
Now, divide both sides of the equation by 4:
$$y = -\frac{1}{4}$$
03
Solve the second equation
To solve the second equation \((2y + 3) = 0\), subtract 3 from both sides of the equation:
\(2y = -3\)
Now, divide both sides of the equation by 2:
$$y = -\frac{3}{2}$$
04
Present the solutions
The solutions for the equation \((4y + 1)(2y + 3) = 0\) are:
$$y = -\frac{1}{4} \quad \text{and} \quad y = -\frac{3}{2}$$
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratic Equations
When faced with a quadratic equation, one of the fundamental methods to find solutions is factoring. To factor a quadratic equation means to break it down into simpler expressions that, when multiplied together, produce the original equation. The typical form of a quadratic equation is \(ax^2 + bx + c = 0\), and the goal is to express it as \( (px + q)(rx + s) = 0 \) where \(p, q, r, \text{and} s\) are real numbers that satisfy the original equation.
Example of Factoring
Consider \( x^2 - 5x + 6 = 0 \). To factor this, we look for two numbers that multiply to 6 (the constant term) and add to -5 (the coefficient of the middle term). Those numbers are -2 and -3, which give us the factors \( (x - 2)(x - 3) = 0 \). Factoring reduces the complexity of solving quadratic equations by transforming them into a product of simpler linear factors, leveraging the zero product property.Zero Product Property
The zero product property is a key concept in algebra which states that if a product of two factors equals zero, then at least one of the factors must be zero. This property is critical when we factor a quadratic equation because it allows us to set each factor equal to zero and solve for the variable.
Application in Equations
Taking the factored form of a quadratic equation, \( (px + q)(rx + s) = 0 \), according to the zero product property, we can set \( px + q = 0 \) and \( rx + s = 0 \) separately. Each equation can then be solved for \( x \), providing the possible solutions for the original quadratic equation. This simplifies the problem to solving two linear equations, which is typically more straightforward.Algebraic Solution Steps
Solving quadratic equations algebraically involves a series of structured steps. Following these steps systematically can make finding the solutions more manageable.
Typical Steps for Solving
- Step 1: Write the quadratic equation in standard form, if not already presented as such.
- Step 2: Factor the equation into two binomials, if possible.
- Step 3: Apply the zero product property by setting each binomial equal to zero.
- Step 4: Solve each resulting linear equation to find the values of the variable.
- Step 5: Verify the solutions by plugging them back into the original equation.