what is the solution to the equation $\frac{1}{x}=\frac{x + 3}{2x^{2}}$?\n$x=-3$\n$x=-3$ and $x = 0$\n$x =…

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

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

Answer

Explanation:

Step1: Cross - multiply

Cross - multiply the given equation $\frac{1}{x}=\frac{x + 3}{2x^{2}}$ to get $2x^{2}=x(x + 3)$.

Step2: Expand the right - hand side

Expand $x(x + 3)$ using the distributive property: $2x^{2}=x^{2}+3x$.

Step3: Move all terms to one side

Subtract $x^{2}$ and $3x$ from both sides: $2x^{2}-x^{2}-3x=0$, which simplifies to $x^{2}-3x = 0$.

Step4: Factor the left - hand side

Factor out an $x$: $x(x - 3)=0$.

Step5: Solve for $x$

Set each factor equal to zero. If $x=0$, the original equation $\frac{1}{x}=\frac{x + 3}{2x^{2}}$ has undefined terms (since division by zero is not allowed). If $x-3=0$, then $x = 3$.

Answer:

D. $x = 3$