Problem 31 - Precalculus : Building Concepts and Connections

Question

Solve the rational equation. Check your solutions. $$\frac{x}{2 x^{2}+x-3}+\frac{1}{x-1}=\frac{3}{2 x+3}$$

Step-by-Step Solution

Answer

The solutions to the given equation are \(x = -1\) and \(x = 1\).

Step 1: Find the LCM of the denominators

The denominators are \(2x^2 + x - 3\), \(x - 1\) and \(2x + 3\). Factoring these three, \(2x^2 + x - 3\) factors to \((2x - 1)(x + 3)\), \(x - 1\) is already a factor, and \(2x + 3\) is also its own factor. Their LCM is therefore: \((2x - 1)(x + 3)(x - 1)\).

Step 2: Clear out the fractions

Multiply each term in the equation by the LCM. This yields the equation: \(x(x + 3) + (2x - 1)(x + 3) = 3(x - 1).\)

Step 3: simplify the equation

Expanding and simplifying the above equation, we get: \(x^2 + 3x + 2x^2 - x - 3 = 3x - 3\). Further simplification gives: \(3x^2 + 2x - 3= 0.\)

Step 4: Solve the resulting quadratic equation

The quadratic equation is \(3x^2 + 2x - 3 = 0\). It can be solved using the Quadratic formula \(x = -b ± \sqrt{b^2 - 4ac}/2a .\) This gives \(x = -1, 1\).

Step 5: Check the solutions

Substitute \(x = -1\) and \(x = 1\) into the original equation. After simplification, both values should satisfy the equation.

Step 6: Final Answer

The solutions to the equation are \(x = -1\) and \(x = 1\).

Key Concepts

LCM (Least Common Multiple)

When dealing with rational equations, often the first step is to find the Least Common Multiple (LCM) of the denominators. This helps to eliminate fractions, which simplifies the equation. Consider the equation:

• \(2x^2 + x - 3\)
• \(x - 1\)
• \(2x + 3\)

Finding the LCM involves identifying the smallest expression that each denominator can divide into without a remainder. Here’s a step-by-step breakdown:

Factor Each Denominator

Start by factoring each denominator to its simplest form. For example:

• \(2x^2 + x - 3\) factors to \((2x - 1)(x + 3)\)
• \(x - 1\) is already a simple factor
• \(2x + 3\) remains as it is

Combine Factors

Next, combine all factors, ensuring each appears once. So, the LCM for these would be \((2x - 1)(x + 3)(x - 1)\). Using this LCM allows you to clear fractions by multiplying each term in the equation, making it simpler to solve.

Quadratic Equation

A quadratic equation is any equation that can be rearranged in standard form as \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants. In our rational equation, after clearing the fractions using the LCM, we end up with such an equation:

Forming the Quadratic

Upon multiplying and simplifying, the equation becomes \(3x^2 + 2x - 3 = 0\). This equation is quadratic because the highest power of \(x\) is 2.

• \(a = 3\)
• \(b = 2\)
• \(c = -3\)

Quadratics like this can have various solving methods, but one effective way is using the Quadratic Formula.

Why Solve Quadratics?

Quadratic equations can reveal important values, like the roots where the equation equals zero. These roots in the context of a rational equation provide the solutions that satisfy the original equation.

Quadratic Formula

The Quadratic Formula is a tool that allows you to find the solutions (roots) of any quadratic equation quickly and efficiently. Its formula is:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}.\]

Using the Formula

For \(3x^2 + 2x - 3 = 0\), you can substitute:

• \(a = 3\)
• \(b = 2\)
• \(c = -3\)

Substitute these into the formula to calculate:

• Discriminant: \(b^2 - 4ac = 2^2 - 4(3)(-3) = 4 + 36 = 40\)
• Roots: \(x = \frac{{-2 \pm \sqrt{40}}}{{6}}\)

Interpreting the Roots

The solutions to the quadratic, calculated as \(x = -1\) and \(x = 1\), represent the x-values that satisfy the original equation. Always substitute back to check for validation, ensuring these solutions don't result in undefined terms in the original rational equation.