Problem 13 - Algebra for College Students

Question

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$3 t^{2}+14 t-5=0$$

Step-by-Step Solution

Answer

The solutions are \(t = \frac{1}{3}\) and \(t = -5\).

Step 1: Identify the type of quadratic equation

The given equation is \(3t^2 + 14t - 5 = 0\). It is a quadratic equation in the standard form \(ax^2 + bx + c = 0\), where \(a = 3\), \(b = 14\), and \(c = -5\).

Step 2: Use the factoring method

We aim to factor the quadratic expression into two binomials, i.e., \((pt + q)(rt + s) = 0\). To do this, we need two numbers whose product is \(a \times c = 3 \times (-5) = -15\) and whose sum is \(b = 14\).

Step 3: Find two numbers that multiply to -15 and add to 14

The two numbers that satisfy these conditions are 15 and -1, since \(15 \times (-1) = -15\) and \(15 + (-1) = 14\).

Step 4: Rewrite the middle term with these numbers

Rewrite \(14t\) as \(15t - t\), so the equation becomes \(3t^2 + 15t - t - 5 = 0\).

Step 5: Factor by grouping

Group the terms: \((3t^2 + 15t) + (-t - 5) = 0\). Factor each group separately: \(3t(t + 5) - 1(t + 5) = 0\).

Step 6: Factor out the common binomial

Notice that \((t + 5)\) is common in both groups, so factor it out: \((3t - 1)(t + 5) = 0\).

Step 7: Set each factor equal to zero and solve for t

We solve \(3t - 1 = 0\) and \(t + 5 = 0\). For \(3t - 1 = 0\), add 1 to both sides to get \(3t = 1\), then divide by 3 to find \(t = \frac{1}{3}\). For \(t + 5 = 0\), subtract 5 from both sides to find \(t = -5\).

Key Concepts

Factoring Techniques

Factoring is a crucial concept in algebra, particularly in solving quadratic equations. It's all about breaking down an expression into simpler terms or factors that, when multiplied together, give us the original expression. For quadratic equations, such as the one we have here, the goal is to rewrite the expression in the form of

• \( (pt + q)(rt + s) = 0 \) .

The task is to identify two numbers: these should multiply to result in the product of the quadratic's leading coefficient and the constant term, in this case, \( 3 \times (-5) = -15 \). Additionally, these same numbers must add up to the middle term coefficient, which is \( 14 \) in this scenario.
These numbers, 15 and -1 here, help break down the middle term for the factorization process. Factoring simplifies the equation and paves the way to solve it efficiently, setting up an equation where the zero product property can be applied.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra. The typical form of such an equation is

• \( ax^2 + bx + c = 0 \).

A popular method to tackle these equations is factoring. Once the quadratic is factored, the equation is expressed as a product of two binomials set to zero. The strength of this method lies in the zero product property. If the product of two numbers is zero, at least one of the numbers must be zero.
This allows us to isolate each binomial equal to zero and solve for the variable. To illustrate:

• if you have \((3t - 1)(t + 5) = 0\),

This translates to solving

• \(3t - 1 = 0\)
• and \(t + 5 = 0\).

By solving these smaller, linear equations, we find the solutions to the original quadratic. These are the values of \(t\) that satisfy the equation – here, \(t = \frac{1}{3}\) and \(t = -5\). Breaking it down into smaller, more manageable parts makes the problem much easier.

Group Factorization

Group factorization is a helpful technique when dealing with complex quadratic expressions. When the equation is difficult to factor straightforwardly, this method can simplify it by grouping terms strategically. This helps us in revealing common factors that can be pulled out, making the process more manageable.Let's take the quadratic

• \( 3t^2 + 14t - 5 \),

which can be rearranged to \( 3t^2 + 15t - t - 5 \). The trick is to rewrite the middle term using numbers determined from factoring techniques (i.e., numbers adding to 14 and multiplying to –15).
When split into

• \( (3t^2 + 15t) + (-t - 5) \),

the equation is dissected into simpler parts:

• \(3t(t + 5) - 1(t + 5) = 0 \).

Notice how grouping has highlighted a common factor of \( (t + 5) \) in both terms. This setup allows it to be factored out with ease, yielding \( (3t - 1)(t + 5) = 0 \). Group factorization thus cleverly simplifies solving complex quadratics by focusing on smaller, factorable expressions within the larger equation. This division lets you manage, factor, and ultimately solve the equation accurately.