Problem 48 - Precalculus Mathematics for Calculus

Question

Solve the equation by factoring. $$2 y^{2}+7 y+3=0$$

Step-by-Step Solution

Answer

The solutions are \( y = -\frac{1}{2} \) and \( y = -3 \).

Step 1: Understand the Equation Structure

The given equation is a quadratic equation in the form of \( ax^2 + bx + c = 0 \) where \( a = 2 \), \( b = 7 \), and \( c = 3 \). The goal is to factor this equation into the form \( (dy + e)(fy + g) = 0 \).

Step 2: Multiply Coefficient of\( y^2 \) and Constant Term

Multiply the coefficient of \( y^2 \) which is 2, by the constant term, which is 3. This gives \( 2 \times 3 = 6 \). Find two numbers that multiply to 6 and add up to the middle coefficient, which is 7.

Step 3: Find Factors of the Middle Term

The numbers that multiply to 6 and also add to 7 are 6 and 1. Rewrite the middle term using these two numbers: \( 7y = 6y + y \). The equation is now \( 2y^2 + 6y + y + 3 = 0 \).

Step 4: Group Terms

Group the terms into two pairs: \((2y^2 + 6y) + (y + 3) = 0 \).

Step 5: Factor by Grouping

Factor out the greatest common factor from each pair: \( 2y(y + 3) + 1(y + 3) = 0 \).

Step 6: Factor Out the Common Binomial

The expression can now be written as \( (2y + 1)(y + 3) = 0 \).

Step 7: Solve the Factored Equation

Set each factor equal to zero and solve for \( y \): 1. \( 2y + 1 = 0 \) leads to \( y = -\frac{1}{2} \). 2. \( y + 3 = 0 \) leads to \( y = -3 \).

Key Concepts

Factoring Quadratic Equations

Factoring quadratic equations is a more accessible approach to solving these types of equations. The process involves expressing the quadratic equation as a product of its factors. When you factor, you are essentially reversing the process of expanding the binomials. This is because the ultimate form of a quadratic equation is

• ax^2 + bx + c = 0.

After factoring, the quadratic should look something like this:

• (dx + e)(fx + g) = 0

At this point, each factor can be set to zero to solve for the variable involved. Factoring is especially useful because it provides a straightforward way to find the equation's roots.

Keep in mind that not all quadratic equations are immediately factorable. Sometimes, factoring requires additional techniques or slight changes like re-grouping terms. Being comfortable with different factoring techniques is crucial in finding a solution efficiently.

Quadratic Equation Structure

The structure of a quadratic equation is fundamentally important to understand. A typical quadratic equation follows this pattern:

• ax^2 + bx + c = 0

Here, the letters a, b, and c represent numerical coefficients, with 'a' being the coefficient of the quadratic term, 'b' the coefficient of the linear term, and 'c' the constant term.

Understanding each part's role helps in knowing how to approach solving it. The choice of methods like factoring truly depends on examining these components. For example, knowing the value of 'a' is crucial because it might hint at the complexity of the required steps to factor the equation.

The given equation in this exercise is

• 2y^2 + 7y + 3 = 0

with a = 2, b = 7, and c = 3. Recognizing this structure will help you arrange your terms in the right order for factoring or other solving methods.

Factoring by Grouping

Factoring by grouping is a method you can use when a quadratic equation doesn't factor neatly into two simple binomials right away. This process involves slightly rearranging and re-grouping terms to make factoring possible.

In the exercise equation, the initial quadratic form doesn't factor neatly at a glance. By multiplying 'a' and 'c', you identify a product whose factors can break down the middle term into two separate terms. In this case, calculating 2 (the coefficient of the quadratic term) times 3 (the constant term) gives 6. You find two numbers that multiply to 6 and add to 7 (the middle term), which are 6 and 1.

You would rewrite the quadratic component as:

• 2y^2 + 6y + y + 3 = 0

By grouping terms like

• (2y^2 + 6y) + (y + 3),

you create two pairs with common factors, which are then factored individually. This results in the factors

• 2y(y + 3) + 1(y + 3) = 0.

With the binomial

• (y + 3)

as a common factor, you then have

• (2y + 1)(y + 3) = 0.

This makes it possible to set each factor equal to zero, leading to the equation's solutions.