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Chapter 8: Problem 9

Solve each equation. Check the solutions. $$1-\frac{3}{x}-\frac{28}{x^{2}}=0$$

Short Answer

Expert verified
The solutions are \(x = 7\) and \(x = -4\).

Step by step solution

01

Identify a common denominator

The given equation is \(1 - \frac{3}{x} - \frac{28}{x^2} = 0\). Identify a common denominator for the terms in the equation. Here, the common denominator is \(x^2\).
02

Multiply through by the common denominator

Multiply every term in the equation by \(x^2\) to clear the denominators: \(x^2(1) - x^2\left(\frac{3}{x}\right) - x^2\left(\frac{28}{x^2}\right) = 0\). Simplify to get: \(x^2 - 3x - 28 = 0\).
03

Solve the quadratic equation

This is a standard quadratic equation in the form \(ax^2 + bx + c = 0\). Factorize \(x^2 - 3x - 28 = 0\). The factors of \(-28\) that add up to \(-3\) are \(-7\) and \(4\). So, the equation can be written as \((x - 7)(x + 4) = 0\).
04

Find the solutions

Set each factor equal to zero and solve for \(x\): \(x - 7 = 0 \implies x = 7\). \(x + 4 = 0 \implies x = -4\).
05

Verify the solutions

Substitute \(x = 7\) and \(x = -4\) back into the original equation to verify: For \(x = 7\): \(1 - \frac{3}{7} - \frac{28}{49} = 1 - \frac{3}{7} - \frac{4}{7} = 1 - 1 = 0\), which is true.For \(x = -4\): \(1 - \frac{3}{-4} - \frac{28}{(-4)^2} = 1 + \frac{3}{4} - \frac{28}{16} = 1 + \frac{3}{4} - \frac{7}{4} = 1 - 1 = 0\), which is also true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Denominator
In algebra, a common denominator is crucial when dealing with equations involving fractions. Think of the common denominator as a tool that allows us to combine or compare fractions more easily. For instance, in the equation \(1 - \frac{3}{x} - \frac{28}{x^2} = 0\), each term has a different denominator.
To tackle this, we identify the common denominator, in this case, \(x^2\).
Multiplying everything by \(x^2\) eliminates the fractions and transforms the equation into a simpler one: \(x^2 - 3x - 28 = 0\). This step is fundamental because it allows us to work with a standard quadratic equation, making the solution process much more straightforward.
Factoring Quadratic Equations
Factoring quadratic equations is like breaking down a complex problem into smaller, more manageable pieces. Consider the quadratic equation \(x^2 - 3x - 28 = 0\).
Our goal here is to rewrite it as a product of two binomials. This helps us find the solutions for \(x\).
First, identify two numbers that multiply to give \(-28\) and add to give \(-3\). These numbers are \(-7\) and \(4\).
Therefore, we can factor the equation as \((x - 7)(x + 4) = 0\).
Next, setting each factor equal to zero gives us the solutions \(x - 7 = 0\) and \(x + 4 = 0\), resulting in \x = 7\ and \x = -4\.
Verifying Solutions
Verifying solutions is a vital step to ensure our answers are correct. Once we've factored \(x^2 - 3x - 28 = 0\) and found \(x = 7\) and \(x = -4\), we need to check if these values satisfy the original equation.
Substitute \(x = 7\): $$1 - \frac{3}{7} - \frac{28}{49} = 0$$ simplifies to $$1 - \frac{3}{7} - \frac{4}{7} = 1 - 1 = 0$$ which holds true.
Similarly, for \(x = -4\), substitute into the original equation: $$1 - \frac{3}{-4} - \frac{28}{16} = 0$$ simplifies to $$1 + \frac{3}{4} - \frac{7}{4} = 1 - 1 = 0$$, which is also true.
By verifying these solutions, we confirm that both \(x = 7\) and \(x = -4\) are indeed correct solutions to the given equation.

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