Problem 39 - Algebra 1

Question

Solve the proportion. Check for extraneous solutions. $$\frac{-2}{q}=\frac{q+1}{q^{2}}$$

Step-by-Step Solution

Answer

The equation does not have a solution because the solution \(q = -1/4\) is extraneous as it doesn't satisfy the original equation.

Step 1: Cross multiply

Cross multiplication is the process of cancelling out denominators within a proportion. The process raises each fraction to the power of another denominator, resulting in an equation without fractions. Implementing this to the given fraction, we get \( -2q^{2} = q(q+1) \)

Step 2: Solve the quadratic equation

Solving the equation involves simplifying the equation and combining like terms. In this case, distribute \(q\) on the right side of the equation as \(q^{2} + q\). A standard quadratic equation is generally expressed as \(ax^{2} + bx + c = 0\) which in this instance would be \(-2q^{2} - q^{2} - q = 0\). Collecting like terms transforms the equation to \(-3q^{2} - q = 0\). The resulting equation can be simplified by factoring out the greatest common factor, here \(q\) , to get \(-q = 3q + 1\)

Step 3: Further simplification and finding the solutions

To determine the roots of the equation,analyze the simplified equation \(-q = 3q + 1\). Add \(q\) to both sides to set the equation to zero, resulting in \(0 = 3q + q + 1\), which simplifies to \(4q + 1 = 0\). To solve for \(q\), subtract 1 from both sides (\(4q = -1\)) and then divide by 4, which gives \(q = -1/4\) as the final solution. Checking for extraneous solutions is done by substituting the solution(s) back into the original equation to determine if they hold true.

Step 4: Check for Extraneous Solutions

Substitute the solution back in the original problem to see if both sides of the equation are equal when \(q = -1/4\). The left side of the equation is \(-2/(-1/4) = 8\). The right side becomes \((-1/4 + 1)/ (-1/4)^2 = 3/16\). The left and right sides are not equal, so this means that \(q = -1/4\) is an extraneous solution.

Key Concepts

Cross Multiplication

Cross multiplication is an efficient technique to resolve proportions, which are equations that set two ratios or fractions equal to each other. This method involves multiplying the numerator of one fraction by the denominator of the other fraction across the equation.
For instance, when you have a proportion like \(\frac{a}{b} = \frac{c}{d}\), cross multiplication would give us \(ad = bc\). In the step-by-step solution provided, we applied cross multiplication to the proportion \(\frac{-2}{q} = \frac{q+1}{q^{2}}\) and obtained \(-2q^{2} = q(q + 1)\). This simplifies the original equation by eliminating the fractions, making it easier to solve for the variable.
It is crucial, however, to be careful when cross multiplying to ensure each step is performed accurately, and that no element of the original proportion is altered in the process.

Quadratic Equations

Quadratic equations form an essential part of algebra and are recognizable by the highest exponent of the variable being two, typically written in the standard form \(ax^2 + bx + c = 0\). The coefficients \(a\), \(b\), and \(c\) reshape the curve's directionality, width, and position on a graph.
To solve these equations, one may employ various techniques such as factoring, completing the square, using the quadratic formula, or graphing. In our exercise, simplification of the cross-multiplied proportion leads to a quadratic equation in the form \(-2q^{2} - q^{2} - q = 0\), which simplifies further to \(-3q^{2} - q = 0\). It is then solved by factoring and following algebraic principles to isolate the variable. Solutions for \(q\), called 'roots' of the equation, represent the points where the graph of the quadratic function intersects the horizontal axis.
Every step in solving a quadratic equation is crucial to finding the correct solution and requires careful examination to ensure accuracy.

Extraneous Solutions

When solving algebraic equations, especially those involving rational expressions or radical signs, we sometimes find solutions that do not satisfy the original equation. These are known as extraneous solutions. They usually arise from the steps in our solution process where we may multiply or square both sides of an equation, potentially introducing solutions that weren't there before.
This is why it's vital to check all potential solutions by plugging them back into the original equation. In our step-by-step solution, after finding \(q = -\frac{1}{4}\), we checked for extraneous solutions and discovered that this value does not hold in the original equation, as it creates inequality when substituted. We spotted this by comparing the results obtained on both sides of the original proportion, clearly finding a mismatch. This vigilance ensures the exclusion of any extraneous solutions and confirms the validity of the actual solutions.

Factoring Quadratic Expressions

Factoring is a technique used to break down quadratic expressions into simpler binomial or trinomial factors, which when multiplied together yield the original quadratic expression. It's a fundamental skill for simplifying and solving quadratic equations without relying on the quadratic formula.
In our exercise, the quadratic equation formed was factored by first finding the greatest common factor, which can often simplify the equation and make the subsequent steps easier. Here, the term \(q\) was factored out, simplifying the equation and setting the stage for finding the roots.
When factoring, it’s important to look for patterns like the difference of squares or perfect square trinomials and to remember that not all quadratic expressions can be factored easily. In some cases, alternative methods such as completing the square or using the quadratic formula may be needed. Factoring provides a direct way to find zeros of a function and hence the solutions to the associated equation.