Problem 71 - Contemporary Precalculus

Question

Solve the equation and check your answers. $$\frac{2}{x^{2}}-\frac{5}{x}=4$$

Step-by-Step Solution

Answer

Question: Solve the equation \(\frac{2}{x^{2}}-\frac{5}{x}=4\) and find the valid solutions for \(x\). Answer: The valid solutions for \(x\) are \(x_1 = \frac{-5 + \sqrt{57}}{8}\) and \(x_2 = \frac{-5 - \sqrt{57}}{8}\).

Step 1: Find Common Denominator

In order to solve the equation, let's first find a common denominator between the two fractions, which in this case is \(x^2\).

Step 2: Rewrite equation with common denominator

Now that we have found the common denominator, let's rewrite the equation using it: $$\frac{2}{x^{2}}-\frac{5(x)}{x(x^{2})}=4$$ $$\frac{2-5x}{x^{2}} = 4$$

Step 3: Clear the denominator

In order to solve the equation, let's multiply both sides by the denominator (\(x^2\)) to clear it: $$(x^2) \left(\frac{2-5x}{x^{2}}\right) = (x^2)(4)$$ $$2 - 5x = 4x^2$$

Step 4: Move all terms to one side to create a quadratic equation to solve

Now we'll have a quadratic equation when we move all the terms to one side of the equation: $$4x^2 + 5x - 2 = 0$$

Step 5: Solve the quadratic equation

To solve the quadratic equation, we can use different methods such as factoring, completing the square, or the quadratic formula. In this case, it is simpler to use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ With \(a = 4\), \(b = 5\), and \(c = -2\), we have: $$x = \frac{-5 \pm \sqrt{5^2 - 4(4)(-2)}}{2(4)} = \frac{-5 \pm \sqrt{57}}{8}$$ So we have two potential solutions: $$x_1 = \frac{-5 + \sqrt{57}}{8}$$ $$x_2 = \frac{-5 - \sqrt{57}}{8}$$

Step 6: Check validity of solutions

Now we must check our answers by plugging them back into the original equation to ensure that they are valid solutions. Checking \(x_1\): \(\frac{2}{\left(\frac{-5 + \sqrt{57}}{8}\right)^{2}} - \frac{5}{\frac{-5 + \sqrt{57}}{8}} = 4\) This holds true, so \(x_1 = \frac{-5 + \sqrt{57}}{8}\) is a valid solution. Checking \(x_2\): \(\frac{2}{\left(\frac{-5 - \sqrt{57}}{8}\right)^{2}} - \frac{5}{\frac{-5 - \sqrt{57}}{8}} = 4\) This also holds true, so \(x_2 = \frac{-5 - \sqrt{57}}{8}\) is a valid solution as well. Thus, both potential solutions are valid: $$x_1 = \frac{-5 + \sqrt{57}}{8}$$ $$x_2 = \frac{-5 - \sqrt{57}}{8}$$

Key Concepts

Solving Equations

Solving equations is like finding the missing piece of a puzzle. Every equation has certain parts, and solving it means figuring out what numbers make the equation true.
In our example, we begin with the equation \(\frac{2}{x^{2}}-\frac{5}{x}=4\). The goal is to find the values of \(x\) that satisfy the equation. This involves several steps, which typically include simplifying the equation, ensuring all parts are manageable.

• Identify what you need to find; here, we look for \(x\).
• Simplify or rearrange if necessary to make calculations easier.
• Use mathematical rules and operations such as subtraction, division, or multiplication to work towards the solution.

Finding the solution often requires a combination of these skills to break down the equation into something more familiar.

Common Denominator

When dealing with fractions, finding a common denominator is crucial. It allows you to combine these fractions easily. In math, a denominator is the bottom number in a fraction, telling us into how many parts the whole is divided.
Finding a common denominator means adjusting fractions so they have the same bottom number. In the given equation, the fractions have different denominators: \(x^{2}\) in \(\frac{2}{x^{2}}\) and \(x\) in \(\frac{5}{x}\).
To simplify, we find a common denominator, which is \(x^{2}\). That means rewriting \(\frac{5}{x}\) as \(\frac{5(x)}{x^{2}}\). This makes it easy to combine the fractions and solve the equation. Remember:

• A common denominator simplifies subtraction or addition of fractions.
• It ensures consistency across the equation, helping further calculations.

Taking this step ensures a smoother path to solving the problem.

Quadratic Formula

The quadratic formula is a universal tool in algebra for solving equations of the form \(ax^2 + bx + c = 0\). It's particularly useful when factoring or other methods are too complex.
In our case, we reach the quadratic equation \(4x^2 + 5x - 2 = 0\). Solving a quadratic equation typically involves:

• Recognizing the coefficients \(a\), \(b\), and \(c\).
• Inserting these into the formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Here, \(a = 4\), \(b = 5\), and \(c = -2\). Plug these numbers into the formula to find \(x\), which gives us the potential solutions:
\(x_1 = \frac{-5 + \sqrt{57}}{8}\) and \(x_2 = \frac{-5 - \sqrt{57}}{8}\).

• The formula is comprehensive, covering all possible cases for quadratic equations.
• It provides the solutions once you determine \(a\), \(b\), and \(c\) correctly.

Checking Solutions

Checking solutions is a critical step to ensure no mistakes were made during calculations. It's like proofreading a paper to catch any errors. After finding potential solutions, plug them back into the original equation to verify they work.
In this problem, we obtained solutions \(x_1\) and \(x_2\) using the quadratic formula. To check:

• Replace \(x\) in the original equation with each calculated solution.
• Calculate to see if the left side equals the right side.

For both solutions, the calculations confirmed they held true, meaning these values of \(x\) make the equation valid.
This step is especially important because:

• It confirms the solutions you found are correct.
• It ensures that all processes followed were accurate, guaranteeing the integrity of your work.

Even small errors can change results, so this step is indispensable in verifying accuracy.