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Chapter 11: Problem 9

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

Short Answer

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

Step by step solution

01

Identify and rewrite the equation

The equation is given as 1 - \frac{3}{x} - \frac{28}{x^2} = 0. To make it easier to solve, let's multiply each term by \(x^2\) to eliminate the denominators.
02

Multiply through by \(x^2\)

Multiply each term by \(x^2\): \(x^2(1) - x^2\left(\frac{3}{x}\right) - x^2\left(\frac{28}{x^2}\right) = 0\). This simplifies to: \(x^2 - 3x - 28 = 0\).
03

Factor the quadratic equation

Now solve the quadratic equation: \(x^2 - 3x - 28 = 0\). We can factor this into: \((x - 7)(x + 4) = 0\).
04

Solve for x

Set each factor equal to zero and solve for \(x\): \(x - 7 = 0\) or \(x + 4 = 0\). Therefore, the solutions are: \(x = 7\) and \(x = -4\).
05

Check the solutions

Substitute \(x = 7\) back into the original equation: \(1 - \frac{3}{7} - \frac{28}{49} = 1 - \frac{3}{7} - \frac{4}{7} = 1 - 1 = 0\), which is true. Substitute \(x = -4\) back into the original equation: \(1 - \frac{3}{-4} - \frac{28}{16} = 1 + \frac{3}{4} - \frac{7}{4} = 1 - 1 = 0\), which is also true. Thus, both solutions are verified.

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Key Concepts

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

Factoring Quadratic Equations
Factoring quadratic equations is an essential skill in algebra. It involves rewriting a quadratic equation of the form \[ ax^2 + bx + c = 0 \]into a product of two binomials. Let鈥檚 take a closer look at how to do this correctly. In our given problem, after multiplying through by the common denominator, we have:\[ x^2 - 3x - 28 = 0 \].To factor this, we need to find two numbers that multiply to \[ c = -28 \]and add up to \[ b = -3 \]. These numbers are -7 and 4. So, we can factor the quadratic equation into:\( (x - 7)(x + 4) = 0 \).Breaking it down this way makes it easier to solve for the values of \( x \).
Checking Solutions
After solving the quadratic equation, it's crucial to check the solutions by substituting them back into the original equation. This ensures they are correct. Let鈥檚 verify our solutions: First, substitute \( x = 7 \) into the original equation:\[ 1 - \frac{3}{7} - \frac{28}{49} = 1 - \frac{3}{7} - \frac{4}{7} = 1 - 1 = 0 \].The left-hand side equals zero, so \( x = 7 \) is a solution. Next, substitute \( x = -4 \):\[ 1 - \frac{3}{-4} - \frac{28}{16} = 1 + \frac{3}{4} - \frac{7}{4} = 1 - 1 = 0 \].This also checks out, confirming \( x = -4 \) as a solution.
Multiplying Through by Common Denominator
Multiplying through by a common denominator is often the first step for handling equations involving fractions. This process helps eliminate fractions, simplifying the equation. In our problem, the original equation was:\( 1 - \frac{3}{x} - \frac{28}{x^2} = 0 \).By multiplying each term by \( x^2 \), the smallest common multiple of the denominators (1, x, x^2), we get:\[ x^2(1) - x^2\frac{3}{x} - x^2\frac{28}{x^2} = x^2 - 3x - 28 = 0 \].This transforms a complex fraction equation into a simpler quadratic equation, making it manageable to solve.

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