Question

what is the solution to the equation \\(\frac{1}{x} = \frac{x + 3}{2x^2}\\)? \\(\bigcirc\\ x = -3\\) \\(\bigcirc\\ x = -3\\) and \\(x = 0\\) \\(\bigcirc\\ x = 0\\) and \\(x = 3\\) \\(\bigcirc\\ x = 3\\)

Explanation:

Step1: Identify restrictions

The original equation is \( \frac{1}{x}=\frac{x + 3}{2x^{2}} \). We know that \( x
eq0 \) because division by zero is undefined. So \( x = 0 \) cannot be a solution, eliminating options with \( x = 0 \).

Step2: Cross - multiply to solve

Cross - multiply the equation \( \frac{1}{x}=\frac{x + 3}{2x^{2}} \) (since \( x
eq0 \), we can do this). We get \( 2x^{2}\times1=x\times(x + 3) \).

Step3: Simplify the equation

Simplify the right - hand side: \( 2x^{2}=x^{2}+3x \).
Subtract \( x^{2}+3x \) from both sides: \( 2x^{2}-x^{2}-3x = 0 \), which simplifies to \( x^{2}-3x = 0 \).
Factor out an \( x \): \( x(x - 3)=0 \).
The solutions to this equation are \( x = 0 \) or \( x=3 \). But we already know from the domain restriction that \( x
eq0 \), so the only valid solution is \( x = 3 \).

Answer:

D. \( x = 3 \) (assuming the last option is D, if options are labeled as A, B, C, D with the last one being \( x = 3 \))