Problem 22 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ 1+\frac{2 x}{(x+3)(x+4)}=\frac{2}{x+3}+\frac{4}{x+4} $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{-3 \pm \sqrt{41}}{2} \).

Step 1: Simplify Right-Hand Side

The right-hand side of the equation is \( \frac{2}{x+3} + \frac{4}{x+4} \). To simplify, we find a common denominator, which is \((x+3)(x+4)\). Rewrite each fraction with this common denominator: \[ \frac{2(x+4) + 4(x+3)}{(x+3)(x+4)} \] Simplifying the numerator gives \( 2x + 8 + 4x + 12 = 6x + 20 \). So, the right-hand side becomes \( \frac{6x + 20}{(x+3)(x+4)} \).

Step 2: Combine Terms on the Left-Hand Side

The left-hand side is \( 1 + \frac{2x}{(x+3)(x+4)} \). Rewrite '1' as \( \frac{(x+3)(x+4)}{(x+3)(x+4)} \) to have a common denominator: \[ \frac{(x+3)(x+4) + 2x}{(x+3)(x+4)} \] The numerator becomes: \( x^2 + 7x + 12 + 2x = x^2 + 9x + 12 \). Thus, the left-hand side is \( \frac{x^2 + 9x + 12}{(x+3)(x+4)} \).

Step 3: Set the Numerators Equal to Each Other

Since both sides of the equation have the same denominator, set the numerators equal: \[ x^2 + 9x + 12 = 6x + 20 \] Simplify to get a quadratic equation: \[ x^2 + 9x + 12 - 6x - 20 = 0 \] Which simplifies to \( x^2 + 3x - 8 = 0 \).

Step 4: Solve the Quadratic Equation

To solve \( x^2 + 3x - 8 = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 3 \), \( c = -8 \). Calculate the discriminant: \( b^2 - 4ac = 3^2 - 4(1)(-8) = 9 + 32 = 41 \). Thus, the solutions are: \( x = \frac{-3 \pm \sqrt{41}}{2} \).

Step 5: Verify the Solutions in the Original Equation

Check that the solutions do not make any denominators zero. The original denominators \((x+3)\) and \((x+4)\) must not evaluate to zero. Since \( x = \frac{-3 \pm \sqrt{41}}{2} \) does not simplify to -3 or -4, both solutions are valid.

Key Concepts

Quadratic Equations

Quadratic equations are a cornerstone of algebra and appear in the form \( ax^2 + bx + c = 0 \). The solution to these equations can be found using different methods like factoring, completing the square, or the quadratic formula.
Quadratic equations arise in various real-world scenarios ranging from physics to finance. In this exercise, we derived our quadratic equation from setting two algebraic expressions equal.
In step 4 of the solution, we applied the quadratic formula. This formula is:

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

Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). By identifying these values from our equation, we can solve for x. The formula effectively finds the points where the quadratic equation crosses the x-axis.

Numerators and Denominators

Fractions are made up of numerators and denominators. The numerator is the top part, representing parts of a whole, while the denominator is the bottom part, indicating the total number of equal parts.
In this exercise, understanding numerators and denominators was crucial for simplifying expressions on both sides of the equation.
We began by rewriting the fractions on each side with a common denominator. This step involved multiplying parts of these fractions to have a uniform denominator, \((x+3)(x+4)\).
When working with complex fractions, finding a common denominator helps to combine or compare them efficiently. It allows us to focus on the numerators since the denominators will cancel out, simplifying the overall process.

Simplifying Expressions

To solve algebraic equations, one must often simplify expressions. Simplification involves combining like terms and reducing fractions or expressions to their simplest form.
In our equation, simplification was achieved by finding a common denominator for fractions on both sides and then combining terms. For instance, on the right side,

• \( \frac{2}{x+3} + \frac{4}{x+4} \)

was simplified to

• \( \frac{6x + 20}{(x+3)(x+4)} \)

Simplifying ensures that the equation is easier to work with, particularly when attempting to set equal the numerators of both sides. This reduced the original equation to a quadratic form which is much simpler to solve.

Real Solutions

The term 'real solutions' refers to the values of the variable that can satisfy the equation without leading to undefined or non-real (imaginary) outcomes.
In our exercise, real solutions were determined after solving the quadratic equation \( x^2 + 3x - 8 = 0 \) using the quadratic formula and ensuring these solutions make sense in the context of the problem.
During the last step, we verified our solutions by checking they do not make any original denominators zero, as such instances would render the solution invalid. Since our calculated solutions \( \frac{-3 \pm \sqrt{41}}{2} \) did not cause the denominators \( (x+3) \) or \( (x+4) \) to become zero, both were indeed valid real solutions.
Continuously ensuring solutions remain within the realm of real numbers is crucial for many algebraic and real-life problem-solving scenarios.