Problem 40 - Intermediate Algebra

Question

Solve each equation. If a solution is extraneous, so indicate. \(\frac{3}{x-4}+\frac{2}{x+5}+\frac{18}{x^{2}+x-20}=0\)

Step-by-Step Solution

Answer

The equation has no real solution; \(x = -5\) is extraneous.

Step 1: Simplify the expression

First, we factorize the quadratic expression in the denominator: \[ x^2 + x - 20 = (x - 4)(x + 5) \]This allows us to write the equation as a single fraction with a common denominator: \[\frac{3(x+5) + 2(x-4) + 18}{(x-4)(x+5)} = 0 \]This simplifies to:\[ \frac{3x + 15 + 2x - 8 + 18}{(x-4)(x+5)} = 0 \] simplifying further gives:\[ \frac{5x + 25}{(x-4)(x+5)} = 0 \]

Step 2: Set the numerator equal to zero

To solve the equation, set the numerator equal to zero:\[ 5x + 25 = 0 \]Solve for \(x\) by subtracting 25 from both sides and then dividing by 5:\[5x = -25 \]\[x = -5 \]

Step 3: Check for extraneous solution

Our solution, \(x = -5\), must be checked against the original equation’s denominators to ensure it does not make any term undefined. Check the conditions:1. \(x - 4 eq 0 \Rightarrow x eq 4 \)2. \(x + 5 eq 0 \Rightarrow x eq -5 \)Since \(x = -5\) makes \(x + 5 = 0\) (which makes a denominator zero), \(x = -5\) is extraneous.

Key Concepts

Extraneous Solutions

In rational equations, extraneous solutions are solutions that arise during the process of solving the equation but do not satisfy the original equation. Such solutions often occur when both sides of the equation are manipulated, such as by multiplying both sides by an expression with variables. This manipulation can introduce solutions that don't actually work in the starting equation.

It's crucial to double-check any potential solution by plugging it back into the original equation. You need to ensure it doesn't cause any denominators to become zero, making the equation undefined. For example, if a solution makes the denominator zero, it is not valid. This step is vital and should never be skipped.

In our exercise, the solution was found to be \(x = -5\), but after checking, we realized it made the denominator \(x + 5 = 0\). Hence, it was rightfully discarded as it leads to an undefined expression.

Common Denominator

When solving rational equations like the one given, finding a common denominator is an important step. A common denominator allows you to combine terms effectively and simplify the equation to a more manageable form. This process involves identifying the least common multiple of all the denominators present in the equation.

In the explored exercise, the denominators are \(x-4\), \(x+5\), and \(x^2+x-20\). The quadratic \(x^2 + x - 20\) can be factorized into \((x-4)(x+5)\), making it immediately apparent that the common denominator for all terms is \((x-4)(x+5)\).

• Factorize any quadratic expressions to reveal potential common factors.
• Ensure that each term of the original rational equation shares this common denominator for successful combination.

With a common denominator, we can then consolidate the terms into a single fraction, simplifying the equation for further calculations.

Quadratic Factorization

Quadratic factorization plays a pivotal role in both simplifying rational equations and identifying common denominators. The method relies on breaking down complex quadratic expressions into simpler binomial expressions, which can be used to simplify equations or solve them.

For example, in the exercise, we factorized \(x^2 + x - 20\) into \((x-4)(x+5)\). This works by finding two numbers that multiply to give the constant term (-20 in this case) and add to give the linear coefficient (1 here).

• Look for pairs of factors of the constant term that sum to the linear coefficient.
• Rewriting the quadratic as a product of two binomial terms allows easier manipulation.

Understanding quadratic factorization not only simplifies the calculations but is critical in achieving the further step of finding a common denominator efficiently. This technique is heavily utilized in algebra to solve problems involving polynomials.