Problem 71 - Algebra for College Students

Question

Solve each of the equations. $$-2 y=4 y^{2}$$

Step-by-Step Solution

Answer

Solutions: y = 0 and y = -2.

Step 1: Write the Equation in Standard Form

Start by moving all terms onto one side of the equation to set the equation to 0. Subtract \(-2y\) from both sides to get the standard form.\[-2y - 4y^2 = 0\]

Step 2: Factor the Equation

Factor out the greatest common factor from the equation. In this case, factor out \(-2y\) from each term.\[-2y(y + 2) = 0\]

Step 3: Solve Each Factor

Setting each factor equal to zero, solve for \(y\).For \(-2y = 0\), divide by \(-2\) on both sides:\[y = 0\]For \(y + 2 = 0\), subtract 2 from both sides:\[y = -2\]

Step 4: Write Down the Solutions

After solving each part, we find the solutions for \(y\). The solutions are \(y = 0\) and \(y = -2\).

Key Concepts

Factoring Polynomials

When we face a quadratic equation such as \(-2y - 4y^2 = 0\), factoring polynomials becomes a crucial strategy to simplify and solve it. Factoring involves expressing a polynomial as the product of its factors, which can make complex expressions easier to understand and solve.

• In our example, we first notice that both terms share common factors, \(-2y\) in this case.
• By factoring out \(-2y\), we transform the expression to \(-2y(y + 2) = 0\). This step is important as it prepares the equation for solving by isolating the variable product.
• Ultimately, factoring transforms a complex equation into simpler, easy parts we can solve individually by setting each factor equal to zero.

Understanding the significance of factoring can immensely benefit the process of solving quadratics and higher-degree polynomials as well.

Solving Equations

After factoring our quadratic equation \(-2y(y + 2) = 0\), the next step involves solving each part separately. This method is known as "finding the roots." The fundamental principle here is that if a product of several terms equals zero, at least one of the terms must be zero.

• For \(-2y = 0\), we isolate \(y\) by dividing both sides by \(-2\), yielding \(y = 0\).
• For \(y + 2 = 0\), we simply subtract 2 from both sides to find \(y = -2\).
• This process gives us multiple solutions: \(y=0\) and \(y=-2\), which are known as the roots or solutions of the equation.

Recognizing this principle and applying these straightforward algebraic manipulations enables us to tackle similar problems with confidence.

Algebraic Manipulation

Algebraic manipulation allows us to transform and solve equations smoothly. This involves standard techniques such as moving terms around, factoring, and performing arithmetic operations like addition, subtraction, multiplication, or division.

• Initially, we adjusted the original equation by moving \(-2y\) to one side to get a standard form equation, \(-2y - 4y^2 = 0\). This rearrangement is a step into solving any equation correctly.
• Then, we factored out the greatest common factor, in this case, \(-2y\), to simplify the equation into \(-2y(y+2)=0\).
• By using these algebraic manipulation techniques, we get to the root of the problem without losing sight of important steps or solutions.

Such manipulations require practice to master and are the cornerstone of efficiently working through algebraic equations.