Problem 28 - A Graphical Approach to College Algebra

Question

Find all complex solutions for each equation by hand. Do not use a calculator. $$\frac{2 x}{x-3}+\frac{4}{x+3}=\frac{24}{9-x^{2}}$$

Step-by-Step Solution

Answer

The only complex solution is \(x = -2\).

Step 1: Identify the Common Denominator

To combine the fractions on the left-hand side and equate with the right-hand side, identify the common denominator for the equation. Notice the denominators: \(x-3\), \(x+3\), and \(9-x^2\). Since \(9-x^2 = (3-x)(3+x)\), rewrite it as \(-(x-3)(x+3)\). The common denominator is \((x-3)(x+3)\).

Step 2: Rewrite Fractions with Common Denominator

Multiply each term by the common denominator \((x-3)(x+3)\) to eliminate the denominators:\[ \frac{2x}{x-3} \times (x-3)(x+3) + \frac{4}{x+3} \times (x-3)(x+3) = \frac{24}{9-x^2} \times (x-3)(x+3) \]Simplifying, this becomes:\[ 2x(x+3) + 4(x-3) = -24 \]

Step 3: Expand and Simplify the Equation

Expand the expressions on the left-hand side:\[ 2x^2 + 6x + 4x - 12 = -24 \]Combine like terms:\[ 2x^2 + 10x - 12 = -24 \]

Step 4: Move All Terms to One Side

To set the quadratic equation to zero: Add 24 to both sides to obtain:\[ 2x^2 + 10x + 12 = 0 \]

Step 5: Simplify and Factor the Quadratic Equation

Divide every term by 2 to simplify:\[ x^2 + 5x + 6 = 0 \]Factor the quadratic as: \[ (x+2)(x+3) = 0 \]

Step 6: Solve for the Roots

Set each factor equal to zero and solve for \(x\):\[ x + 2 = 0 \quad \rightarrow \quad x = -2 \]\[ x + 3 = 0 \quad \rightarrow \quad x = -3 \]

Step 7: Check for Restrictions

Both \(x = -2\) and \(x = -3\) must satisfy the original equation, but note that \(x = -3\) causes a division by zero in the original equation as it results in zero in the denominator \((x+3)\).Thus, the solution \(x = -3\) is redundant.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2. It generally takes the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants with \(a eq 0\) because if \(a=0\), then the equation becomes linear and not quadratic. In our exercise, by moving terms around and simplifying, we eventually reached the quadratic equation \(x^2 + 5x + 6 = 0\). Quadratic equations can have two solutions, one solution, or no real solution at all. The solution to a quadratic can often be found by factoring, completing the square, or using the quadratic formula. However, in many cases, such as this one, using factoring provides a simpler path.

Common Denominator

Finding a common denominator is a key step in simplifying equations that involve fractions. A common denominator is basically the least common multiple of the denominators of the fractions. It allows us to combine or compare the fractions more easily by rewriting them with the same denominator. In our exercise, we had the denominators \(x-3\), \(x+3\), and \(9-x^2\). Noticing that \(9-x^2\) can be rewritten using difference of squares as \(-(x-3)(x+3)\), helps us recognize that the common denominator is \((x-3)(x+3)\). By multiplying each fraction by this common denominator, we eliminate the fractions and simplify further manipulation of the equation.

Factoring

Factoring is breaking down an expression into its simplest components that, when multiplied together, give the original expression. It is a critical skill in algebra for simplifying expressions and solving equations. Once we simplified our quadratic equation to \(x^2 + 5x + 6 = 0\), we looked for two numbers that multiply to 6 (the constant term) and add up to 5 (the linear coefficient). These numbers were 2 and 3, so we factored the quadratic as \((x+2)(x+3) = 0\). Factoring allows us to set each factor equal to zero, providing a straightforward path to finding the solution.

Roots of Equations

The roots of an equation are the solutions where the equation equals zero. For a quadratic equation like ours, the term "roots" refers to the points where the graph of the equation crosses the x-axis. After factoring \((x+2)(x+3) = 0\), we set each factor equal to zero: \(x + 2 = 0\) and \(x + 3 = 0\). Solving these, we find the roots \(x = -2\) and \(x = -3\). However, checking these roots against the original equation is crucial because they must satisfy it. In our case, \(x = -3\) would make one denominator zero in the original equation, so it's not a valid solution. Always check to ensure the roots do not violate any initial conditions set by the problem.