Problem 79 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{3}{s-2}+\frac{s-14}{2 s^{2}-3 s-2}-\frac{4}{2 s+1}=0\)

Step-by-Step Solution

Answer

The solution is \(s = 1\); it is not extraneous.

Step 1: Factor the Denominators

The first step is to factor the denominators. For \(2s^2 - 3s - 2\), the factored form is \((s-2)(2s+1)\). This allows us to write the original equation with common denominators in mind: \(\frac{3}{s-2} + \frac{s-14}{(s-2)(2s+1)} - \frac{4}{2s+1} = 0\).

Step 2: Find a Common Denominator

The common denominator for all terms is \((s-2)(2s+1)\). Rewrite each fraction with this common denominator:\[\frac{3(2s+1)}{(s-2)(2s+1)} + \frac{s-14}{(s-2)(2s+1)} - \frac{4(s-2)}{(s-2)(2s+1)} = 0\]

Step 3: Combine the Fractions

Combine the fractions over the common denominator:\[\frac{3(2s+1) + (s-14) - 4(s-2)}{(s-2)(2s+1)} = 0\] Simplify the numerator separately as follows.

Step 4: Simplify the Numerator

Distribute and combine like terms in the numerator:- \(3(2s+1) = 6s + 3\)- \(4(s-2) = 4s - 8\)Combine: \[6s + 3 + s - 14 - 4s + 8 = 0\]This simplifies to \(3s - 3 = 0\).

Step 5: Solve the Equation

Set the simplified numerator to zero since a fraction is zero when its numerator is zero. Solve for \(s\):\[3s - 3 = 0\]Add 3 to both sides:\[3s = 3\]Divide by 3:\[s = 1\]

Step 6: Check for Extraneous Solutions

Check if \(s = 1\) results in a zero denominator in the original equation. The denominators are \(s-2\), \((s-2)(2s+1)\), and \(2s+1\). For \(s=1\), none of these are zero. Thus, \(s=1\) is not an extraneous solution.

Key Concepts

Rational Expressions

Rational expressions are fractions where the numerator and/or the denominator contain variables, usually within polynomial expressions. They are similar to regular fractions but require special attention since variables can be unknown or undefined at certain values. To solve problems with rational expressions:

• Identify any restrictions by finding values that make the denominator zero.
• Simplify the expressions by factoring and reducing them, if possible.
• Combine like terms and simplify further to ease calculations.

To deal with rational expressions effectively, always consider the restriction that the denominator should never be zero as this would make the expression undefined.

Solving Equations

When solving equations involving rational expressions, the key is to first find a common denominator among the terms. This allows you to combine them into a single rational expression, making it easier to manage.

• Factor divisible polynomials to identify a common denominator.
• Re-express each fraction using this common denominator, allowing you to combine the terms.
• Once combined, solve by equating the simplified numerator to zero.

The method works because a fraction is zero if and only if its numerator is zero, given that the denominator is non-zero. Always ensure that none of the denominators equates to zero for the solution obtained.

Factoring Polynomials

Factoring polynomials is an essential skill in solving rational equations. This involves breaking down a complex polynomial into simpler terms or products that multiply back to the original expression. For instance, factoring a quadratic polynomial like \(2s^2 - 3s - 2\) converts it into

• (s-2)(2s+1), a product of two simpler factors.

This step simplifies the equation significantly, making it easier to manage rational expressions. Factoring breaks down the problem into manageable blocks, allowing for a more straightforward path to finding a common denominator and solving the equation.

Extraneous Solutions

In solving equations involving rational expressions, some solutions might appear valid but are not because they make a denominator zero. These are known as extraneous solutions and occur as a result of operations like squaring both sides or introducing additional terms.

• Always verify each solution against the original equation.
• Look specifically for values that make any of the original denominators zero.

In the provided exercise, the solution \(s=1\) was tested for extraneous results by checking against the original denominators: \(s-2\), \((s-2)(2s+1)\), and \(2s+1\). None were zero at \(s=1\), confirming it as a valid solution. Eliminating extraneous solutions ensures that the results are accurate and applicable to real-world contexts.