Problem 57 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Let \(a\) be a nonzero real number. Tell whether each statement is always true, sometimes true, or never true. Explain your reasoning. a. For the equation \(\frac{1}{x-a}=\frac{x}{x-a}, x=a\) is an extraneous solution. b. The equation \(\frac{3}{x-a}=\frac{x}{x-a}\) has exactly one solution. c. The equation \(\frac{1}{x-a}=\frac{2}{x+a}+\frac{2 a}{x^2-a^2}\) has no solution.

Step-by-Step Solution

Answer

Statement a is sometimes true; specific when \(a=1\). Statement b is always true; it always has one solution. Statement c is never true; as the equation will always have solutions unless discriminant is negative.

Step 1: Statement a Analysis

The given equation is \(\frac{1}{x-a}=\frac{x}{x-a}\). If we equate the two expressions, we simplify it to \(1=x\). Now, let's see what happens when \(x=a\). The equation evaluates to \(1=a\), implying that \(a=1\). It seems \(x=a\) is not extraneous in general but needs \(a=1\). Hence, the equation statement is sometimes true, specifically when \(a=1\).

Step 2: Statement b Analysis

The given equation is \(\frac{3}{x-a}=\frac{x}{x-a}\). Simplifying, we get \(3=x\). It has precisely one solution, and that is when \(x=3\). Therefore, the statement is always true, as no matter what nonzero real number we choose for \(a\), the equation will always have one solution.

Step 3: Statement c Analysis

The given equation here is \(\frac{1}{x-a}=\frac{2}{x+a}+\frac{2 a}{x^2-a^2}\). Simplifying this, we get \(x^2-a^2+2ax-2a=0\). This equation always has a solution, as it is a quadratic equation. So, the statement that this equation has no solution is never true. The equation will always have a solution unless discriminant of the quadratic formula is negative.

Key Concepts

Extraneous Solutions

Extraneous solutions are solutions that you find while solving an equation but don't actually satisfy the original equation. These can occur especially when dealing with rational equations and are introduced when you manipulate the equation, for example by multiplying both sides by a variable term.
When solving equations like \(\frac{1}{x-a} = \frac{x}{x-a}\), the step where we multiply both sides by \((x-a)\) to cancel out the fractions could introduce such solutions.
After simplifying, you might end up with \(1 = x\), which seems legitimate unless \(x = a\) because this makes the original equation undefined.

• Always check potential solutions back in the original equation.
• Verify if any solution makes a denominator zero, thus being extraneous.

It's crucial to identify these extraneous solutions to ensure you don't consider them as valid solutions.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2 and have the general form \(ax^2 + bx + c = 0\). These equations can typically have two solutions, found using the quadratic formula or by factoring, when possible.
When dealing with the equation \(\frac{1}{x-a} = \frac{2}{x+a} + \frac{2a}{x^2-a^2}\), you might initially believe there are no solutions, but this turns into a quadratic equation like \(x^2 - a^2 + 2ax - 2a = 0\).

• Check to see if the equation can be factored.
• Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), if necessary.
• Analyzing the discriminant \(b^2 - 4ac\) can tell you about the nature of solutions; a positive discriminant means two solutions, zero means one real solution, and negative means no real solutions.

Thus, it's essential to understand that even confusing equations often resolve into standard quadratic forms.

Equation Solving Techniques

Solving rational equations often requires employing specific techniques to simplify and resolve them effectively. Choosing the right technique is crucial to avoid errors and ensure a clear path to the solution.
In exercises where an equation like \(\frac{3}{x-a} = \frac{x}{x-a}\) is present, simplifying by reducing fractions or multiplying each side by a common denominator, can help clear the fractions.

• Simplify equations by eliminating fractions early on.
• Use factoring, substitution, or the quadratic formula to solve simplified forms.
• Always watch out for conditions that make parts of the equation undefined, such as denominators turning zero.
• Cross-multiplication often helps but be cautious as it can introduce extraneous solutions.

With careful application of these techniques, working through rational equations becomes much less daunting.