Question

Is it possible to write a rational equation that has the following number of solutions? Justify your answers. a. no solution b. exactly one solution c. exactly two solutions d. infinitely many solutions

Short answer

Yes, it is possible to write a rational equation with a) No solution: If the numerator and denominator equates to zero simultaneously. b) Exactly one solution: The numerator is a linear equation and denominator is a non-zero constant. c) Exactly two solutions: The numerator is a quadratic equation and the roots of this equation do not make the denominator zero. d) Infinitely many solutions: The numerator and denominator are the same non-zero polynomials.

Step-by-step solution

Understanding No Solution Scenario

A rational equation has no solutions if its numerator and denominator equates to zero simultaneously. If the denominator is zero regardless of the numerator's value, the equation is undefined, thus no solution.

Understanding One Solution Scenario

A rational equation has exactly one solution when the numerator equals zero but the denominator does not equate to zero simultaneously. This could happen if the numerator is a linear equation and denominator is a non-zero constant.

Understanding Two Solution Scenarios

A rational equation would have exactly two solutions if the numerator is a quadratic equation and the roots of this equation don't make the denominator zero. By solving the quadratic equation, we will get two solutions.

Understanding Infinitely Many Solutions Scenario

In the case of rational equations, having infinite many solutions implies that the equation is always true regardless of the variable's value. Therefore, this is possible if the numerator and denominator are the same non-zero polynomials.

Key concepts

No Solution in Rational Equations

When students encounter rational equations, a common question is about the potential for a 'no solution' scenario. This can occur when no value of the variable makes the equation true. For rational equations, this type of situation typically happens if the equation is structured such that the denominator becomes zero, since division by zero is undefined in mathematics.

For instance, consider the rational equation \( \frac{1}{x-2} = \frac{3}{x-2} \). If the variable was to take the value of 2, both the numerator and the denominator would simultaneously equate to zero. Since dividing any number by zero cannot yield a definitive result, we say the equation has 'no solution'. A rational equation with this characteristic is sometimes called an inconsistent equation.

It's crucial for students to recognize that an undefined denominator disqualifies all potential solutions and thus the equation is unsolvable within the realm of real numbers.

One Solution in Rational Equations

Rational equations with 'one solution' are quite satisfying to solve because it means you can find a unique value for the variable that satisfies the equation. Typically, this happens when the numerator can be zero while the denominator is non-zero at a certain value of the variable. An easy example to illustrate this is \( \frac{x}{5} = 0 \). Here, it's evident that the only solution is when x = 0, which makes the numerator zero without affecting the denominator.

However, it gets a bit more complex when dealing with polynomials. Take the equation \( \frac{x-1}{x+2} = 0 \). In this case, x = 1 is the one and only solution, since this makes the numerator zero and the denominator remains non-zero (\(3\) in this case). Students should note that when encountering a linear equation in the numerator, there's typically only one solution since a line crosses the x-axis only once.

Two Solutions in Rational Equations

Rational equations that yield 'two solutions' can pose a slightly more advanced challenge. These equations usually involve a quadratic expression in the numerator. For example, consider the equation \( \frac{x^2-5x+6}{x+1} = 0 \). Here, the quadratic equation \(x^2-5x+6\) can be factored to \((x-2)(x-3)\), which gives us two possible solutions for x, namely 2 and 3.

However, the twist is in ensuring that these solutions don't make the denominator zero. For the above example, neither 2 nor 3 makes the denominator, \(x+1\), zero, hence they are valid solutions. Students must be attentive to this detail: each root of the numerator must be examined to determine if it results in a zero denominator. If it does, then that root must be excluded from the set of solutions.

Infinitely Many Solutions in Rational Equations

Encountering a rational equation with 'infinitely many solutions' could initially seem perplexing, but such instances are grounded in a very logical principle. For these equations, every real number is a solution. This can only happen if the numerator and the denominator are equivalent (after simplification), non-zero expressions.

For example, the equation \( \frac{3(x-1)}{3(x-1)} = 1 \) simplifies to \(1=1\), which is always true for any value of x (except where the expression is undefined, which in this case would be x = 1). So, we can ignore that one point where the equation is undefined and state that every other value of x is a solution. Students should remember, however, to always check for points where the equation might be undefined. Equations of this nature embody the idea of identity in algebra — an equality that is valid for all permissible values of the variable.