Problem 81 - Algebra for College Students

Question

Solve each equation for the indicated variable. \(5 b x^{2}-3 a x=0\) for \(x \quad\) 82. \(a x^{2}+b x=0\) for \(x\)

Step-by-Step Solution

Answer

First equation: \(x = 0\) or \(x = \frac{3a}{5b}\). Second equation: \(x = 0\) or \(x = -\frac{b}{a}\).

Step 1: Factor the Common Term (First Equation)

The given equation is \(5bx^2 - 3ax = 0\). First, notice that \(x\) is a common factor in both terms. Factor it out to get: \(x(5bx - 3a) = 0\).

Step 2: Apply the Zero Product Property (First Equation)

Since we have \(x(5bx - 3a) = 0\), according to the zero product property, if a product of two factors is zero, at least one of the factors must be zero. So, set each factor equal to zero: 1. \(x = 0\) 2. \(5bx - 3a = 0\).

Step 3: Solve for x in the Second Equation (First Equation)

For \(5bx - 3a = 0\), solve for \(x\) by isolating it: Add \(3a\) to both sides: \(5bx = 3a\) Divide by \(5b\): \(x = \frac{3a}{5b}\).

Step 4: Solve for x in the First Part of the First Equation

From Step 2, remember that one solution for \(x\) is already \(x = 0\). Thus, the solutions are \(x = 0\) or \(x = \frac{3a}{5b}\).

Step 5: Factor the Common Term (Second Equation)

The given equation is \(ax^2 + bx = 0\). Again, notice that \(x\) is a common factor in both terms. Factor it out to get: \(x(ax + b) = 0\).

Step 6: Apply the Zero Product Property (Second Equation)

Since we have \(x(ax + b) = 0\), apply the zero product property by setting each factor equal to zero: 1. \(x = 0\) 2. \(ax + b = 0\).

Step 7: Solve for x in the Second Equation

For \(ax + b = 0\), solve for \(x\) by isolating it: Subtract \(b\) from both sides: \(ax = -b\) Divide by \(a\): \(x = -\frac{b}{a}\).

Step 8: List the Solutions for the Second Equation

The solutions for the second equation are \(x = 0\) or \(x = -\frac{b}{a}\).

Key Concepts

Zero Product Property

The Zero Product Property is a crucial concept in solving quadratic equations. It states that if the product of two factors is zero, then at least one of the factors must be zero. This principle is fundamental when dealing with quadratic equations because it allows us to easily solve for the variable by setting each factor to zero.
For example, in the equations given in the exercise, after factoring, they result in forms such as \(x(5bx - 3a) = 0\) and \(x(ax + b) = 0\). Applying the zero product property means we consider each factor separately:

• Set the first factor \(x = 0\) and solve; this often gives a straightforward solution.
• Set the second factor, like \(5bx - 3a = 0\), and solve for the variable \(x\).

This property is incredibly useful as it simplifies the solving process by breaking down the equation into simpler, manageable parts.

Factoring Equations

Factoring is the method of expressing a mathematical expression as a product of its factors. In the context of quadratic equations, factoring involves finding two or more expressions whose product gives the original equation.
When given an equation like \(5bx^2 - 3ax = 0\), the first step is to find the greatest common factor, which in this case is \(x\). Factoring it out simplifies the equation to \(x(5bx - 3a) = 0\). Similarly, for \(ax^2 + bx = 0\), factoring out \(x\) gives \(x(ax + b) = 0\).

• Check for a common factor in each term of the quadratic equation.
• Extract the common factor to simplify the expression.

Factoring is essential because it prepares the equation for the application of the zero product property, thus paving the way for finding the solutions.

Solutions of Quadratic Equations

Solutions of quadratic equations are the values of the variable \(x\) that satisfy the equation. In simple cases like the exercise provided, solutions are derived after factoring and applying the zero product property.
Here’s how you find the solutions:

• After factoring, apply the zero product property to set each factor to zero.
• For example, \(x(5bx - 3a) = 0\) leads to two solutions: \(x = 0\) and \(5bx - 3a = 0\).
• Solve \(5bx - 3a = 0\) by isolating \(x\): \(x = \frac{3a}{5b}\).

Hence, the solutions are \(x = 0\) or \(x = \frac{3a}{5b}\) for the first equation. Similarly, for the second equation \(x = 0\) or \(x = -\frac{b}{a}\). These solutions satisfy their respective equations, demonstrating that factoring and using the zero product property can effectively solve quadratic equations.