Problem 28 - Algebra for College Students

Question

Solve each equation. $$ \frac{x}{2 x^{2}+5 x}-\frac{x}{2 x^{2}+7 x+5}=\frac{2}{x^{2}+x} $$

Step-by-Step Solution

Answer

The solutions are \( x = -4 + \sqrt{6} \) and \( x = -4 - \sqrt{6} \).

Step 1: Factorize Denominators

Factorize all the denominators. For \( 2x^2 + 5x \), factor out common factor:\[ 2x^2 + 5x = x(2x + 5) \]For \( 2x^2 + 7x + 5 \), factor by grouping: \[ 2x^2 + 7x + 5 = (2x + 5)(x + 1) \]For \( x^2 + x \), factor out common factor:\[ x^2 + x = x(x + 1) \]

Step 2: Find Common Denominator

The common denominator is the combination of all distinct factors from each expression:\[ x(2x + 5)(x + 1) \]

Step 3: Rewrite Each Term with Common Denominator

Rewrite each fraction with the common denominator:- \( \frac{x}{2x^2 + 5x} = \frac{x(x + 1)}{x(2x + 5)(x + 1)} \)- \( \frac{x}{2x^2 + 7x + 5} = \frac{x(2x + 5)}{x(2x + 5)(x + 1)} \)- \( \frac{2}{x^2 + x} = \frac{2(2x + 5)}{x(2x + 5)(x + 1)} \)

Step 4: Combine the Left Side

Combine the fractions on the left side:\[ \frac{x(x + 1) - x(2x + 5)}{x(2x + 5)(x + 1)} = \frac{x^2 + x - (2x^2 + 5x)}{x(2x + 5)(x + 1)} \]Simplify the numerator:\[ = \frac{x^2 + x - 2x^2 - 5x}{x(2x + 5)(x + 1)} = \frac{-x^2 - 4x}{x(2x + 5)(x + 1)} \]

Step 5: Set Equal to Right Side and Simplify

Set the simplified left side equal to the right side:\[ \frac{-x^2 - 4x}{x(2x + 5)(x + 1)} = \frac{2(2x + 5)}{x(2x + 5)(x + 1)} \]Cross multiply and simplify:\[ -x^2 - 4x = 2(2x + 5) \]Expand the right side:\[ -x^2 - 4x = 4x + 10 \]Rearrange the terms to solve for \( x \): \[ -x^2 - 8x - 10 = 0 \]

Step 6: Solve the Quadratic Equation

Use the quadratic formula to solve \( -x^2 - 8x - 10 = 0 \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = -1, b = -8, c = -10 \).Calculate discriminant:\[ b^2 - 4ac = (-8)^2 - 4(-1)(-10) = 64 - 40 = 24 \]\[ x = \frac{-(-8) \pm \sqrt{24}}{2(-1)} = \frac{8 \pm 2\sqrt{6}}{-2} \]\[ x = -4 \pm \sqrt{6} \]

Step 7: Check for Valid Solutions

Check if both solutions are valid by ensuring the original denominators remain non-zero:For \( x = -4 + \sqrt{6} \) and \( x = -4 - \sqrt{6} \), check each term in the original equation. Since none of the denominators become zero, both are valid solutions.

Key Concepts

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler components called factors, which when multiplied together give the original polynomial. It simplifies problems and helps solve equations. For instance, in our equation, the polynomial \( 2x^2 + 5x \) is factored by taking out the common factor \( x \):

• \( 2x^2 + 5x = x(2x + 5) \)

Another polynomial, \( 2x^2 + 7x + 5 \), is factored by grouping:

• Split into parts: \( (2x^2 + 5x) + (2x + 5) \)
• Factor the pairs: \( (2x + 5)(x + 1) \)

This transformation is crucial as it reveals the fundamental building blocks of the polynomial, making further operations like finding a common denominator easier. Factoring often requires recognizing patterns, taking out common factors, or using special formulas for quadratics.

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions of quadratic equations that take the standard form \( ax^2 + bx + c = 0 \). The formula

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

enables us to calculate exactly where a quadratic function equals zero. In our example, solving the equation

• \( -x^2 - 8x - 10 = 0 \)

nets us the following values for \( a, b, \) and \( c \):

• \( a = -1, b = -8, c = -10 \)

We start by calculating the discriminant, the part under the square root:

• Discriminant \( = 64 - 40 = 24 \)

Finally, plug into the formula to find the solutions

• \( x = -4 \pm \sqrt{6} \)

By providing the roots (or solutions) of the equation, the quadratic formula helps solve equations that are not easily factorable.

Common Denominators

In solving rational equations, rewriting each fraction with a common denominator is a crucial step. This allows us to combine or compare fractions more easily. For our problem, we saw different expressions with varying denominators:

• \( x(2x + 5) \)
• \( (2x + 5)(x + 1) \)
• \( x(x + 1) \)

To combine all these, we need a common denominator that incorporates all distinct factors from each denominator. Thus, the common denominator becomes:

• \( x(2x + 5)(x + 1) \)

Every fraction in the equation is then rewritten over this common denominator. By doing so, the equation is standardized, making it easier to manipulate and solve. This technique is essential when dealing with equations involving fractions, ensuring all parts can be combined correctly.

Cross Multiplication

Cross multiplication is a technique used to solve equations involving two fractions set to each other. It simplifies finding \( x \) by eliminating the denominators through

• multiplication across the equal sign.

In our context, once fractions have a common denominator and the equation is balanced on both sides, we move to:

• \( \frac{-x^2 - 4x}{x(2x + 5)(x + 1)} = \frac{2(2x + 5)}{x(2x + 5)(x + 1)} \)

This equation converts to finding a solution for:

• \( -x^2 - 4x = 2(2x + 5) \)

Using cross multiplication, multiply across the fractions to remove the denominators, and you have a simpler equation. This change transforms a complex equation into a more straightforward polynomial equation that can be solved using methods like the quadratic formula. Cross multiplication streamlines the solution process, making it very handy in clearing fractions.