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Chapter 6: Problem 21

Solve. If no solution exists, state this. $$ \frac{9}{x}=\frac{x}{4} $$

Short Answer

Expert verified
The solutions are \( x = 6 \) and \( x = -6 \).

Step by step solution

01

Cross-multiply the fractions

To eliminate the fractions, cross-multiply the terms. Multiply the numerator of the first fraction by the denominator of the second fraction and the denominator of the first fraction by the numerator of the second fraction: \[ 9 \times 4 = x \times x \]
02

Simplify the equation

Simplify both sides of the equation: \[ 36 = x^2 \]
03

Solve for x

To solve for \( x \), take the square root of both sides of the equation: \[ x = \sqrt{36} \]This gives us two possible solutions: \[ x = 6 \]and \[ x = -6 \]
04

Check the solutions

To ensure both solutions are valid, substitute \( x = 6 \) and \( x = -6 \) back into the original equation and verify: For \( x = 6 \):\[ \frac{9}{6} = \frac{6}{4} \]Both sides simplify to \( 1.5 \), so \( x = 6 \) is a solution.For \( x = -6 \):\[ \frac{9}{-6} = \frac{-6}{4} \]Both sides simplify to \( -1.5 \), so \( x = -6 \) is also a solution.

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Key Concepts

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

Cross-Multiplication
When solving rational equations, cross-multiplication is a powerful technique to eliminate the fractions and simplify the equation. Starting with the equation \(\frac{9}{x} = \frac{x}{4}\), cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other. This essentially 'flattens' the fractions into a more manageable form: \(\text{9} \times \text{4} = \text{x} \times \text{x}\). This step is crucial because it paves the way to simplify the equation further. By executing cross-multiplication correctly, we set the foundation for solving for x.
Simplify Equations
After cross-multiplying, our equation transforms into \(36 = x^2\). Simplifying equations often involves breaking down more complex expressions into simpler building blocks. In this step, both sides of the equation are presented as straightforward values. Since 36 is already simplified, we recognize that \(x^2\) needs to be isolated. Therefore, simplifying means handling the equation in a way that brings forth clarity and guides us closer to our solution. This simplification is a vital step in the problem-solving process and prepares us for solving for x.
Square Root
To solve for x, we take the square root of both sides of the equation \(36 = x^2\). Taking the square root is a method to 'undo' the squaring of x, revealing possible solutions for x. Specifically, \(x = \sqrt{36}\) presents us with two solutions: \(x = 6\) and \(x = -6\). It is essential to consider both the positive and negative roots when dealing with squares, as both values satisfy the equation's requirements. Evaluating these roots and substituting them back into the original equation \(\frac{9}{6} = \frac{6}{4}\) and \(\frac{9}{-6} = \frac{-6}{4}\) reassures us that both x = 6 and x = -6 are indeed valid solutions.

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Most popular questions from this chapter

Perform the indicated operations and simplify. $$ \frac{6 t^{2}-26 t+30}{8 t^{2}-15 t-21} \cdot \frac{5 t+15}{t^{2}-4} \div \frac{5 t+15}{t^{2}-4} $$

Simplify. Do not use negative exponents in the answer. $$ -3^{-2} $$

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