Question

Solve the equation by multiplying each side by the least common denominator. $$\frac{1}{x-4}+1=-\frac{7}{x^{2}+x-20}$$

Short answer

The solution to the equation is \(x = -1\). Note that while \(-5\) solves the intermediate equation, it causes a division by zero in the original equation and therefore is not a valid solution.

Step-by-step solution

Factor the denominators

Factor the denominators to find the least common denominator (LCD). For \(x^2+x-20\) it factors to \((x-4)(x+5)\). So, the LCD is \(x(x-4)(x+5)\).

Multiply each side by the LCD

Multiply each term on both sides of the equation by the LCD, yielding: \[x(x+5)+x(x-4)(x+5)=-7x.\] Expand the terms and simplify the equation to get \(x^2 + 6x + 5= -7x.\)

Collect terms and solve for x

Collect similar terms on one side of the equation, which yields, \(x^2 + 13x + 5 = 0\). This is a quadratic equation and can be solved by factoring, completing the square, or using the quadratic formula. Solving the equation using the quadratic formula or by factoring, gives two possible solutions for x, which are \(-1\) and \(-5\). However, substituting these values back into the original equation, we find that \(-5\) is not a valid solution because it causes the denominator to become zero, resulting in undefined values. Hence, the only solution is \(x = -1\).

Key concepts

Understanding Least Common Denominator

Finding the least common denominator (LCD) is crucial in solving equations involving fractions, as it allows us to combine the fractions into a single unified term. The LCD is the smallest number that each denominator in a set of fractions can divide into without leaving a remainder.

In the context of rational equations, the LCD is not only the 'smallest' in the conventional sense but is the least common multiple of the denominators. To find it, we break each denominator into its prime factors. For example, if our denominators are \(x-4\) and \(x^2+x-20\), we factor them to \(x-4\) and \(x+5)(x-4)\) correspondingly. The LCD will be the product of the distinct factors, thus in this case, \(x(x-4)(x+5)\).

By multiplying each term in the equation by the LCD, we 'clear' the denominators, which simplifies our equation to a polynomial that is easier to solve. This step is vital; it transforms a fractional problem into a regular algebraic equation.

Factoring Quadratic Equations

When a quadratic equation appears in the form of \(ax^2 + bx + c = 0\), factoring is one of the tools we can use to solve it. Factoring refers to breaking down a complex expression into a product of simpler factors. The goal is to rewrite the quadratic as \( (x-n)(x-m) = 0\) where \(n\) and \(m\) are the roots (solutions) of the equation.

Factoring often requires identifying two numbers that multiply to give the product of the coefficient \(a\) and constant term \(c\), and add to give the middle coefficient \(b\). This can be challenging, and sometimes impossible, depending on the numbers involved. If an equation cannot be easily factored using integer coefficients, it may still be factorable using other methods or may require a different approach to solve, such as the quadratic formula or completing the square.

In some cases, like \(x^2 + 13x + 5 = 0\), factoring could be too tough without the assistance of a calculator, so it's important to be equipped with other methods of solving quadratics.

Using the Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations that are difficult or impossible to factor easily. The formula is \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation in the form \(ax^2 + bx + c = 0\).

To use the quadratic formula, it's essential to accurately identify the coefficients and plug them into the formula. Once you have the values plugged in, you carry out the operations indicated: compute the discriminant \(b^2 - 4ac\), take its square root, and then solve for the two possible values of \(x\) by addition and subtraction.

Remember:

• If the discriminant is positive, there are two real solutions.
• If it's zero, there is one real solution.
• If it's negative, the solutions are complex or imaginary.

By using this approach, you can find solutions that factoring might not reveal, as it does not rely on the factors being nice and tidy integers. When you retrieve the solutions, it's always important to check them in the original equation to ensure they don't result in undefined expressions. This checking step is what helps us discard extraneous solutions like the \(x = -5\) in this exercise's context.