What are the solutions to this equation?
$\frac{3}{2 x-5}=x$
A. $x=-3$ and $x=0.5$
B. $x=-2$ and $x=2$
C. $x=-0.5$ and $x=3$
D. $x=-1$ and $x=-0.4$
Answer (1)
Multiply both sides by ( 2 x − 5 ) to get 3 = x ( 2 x − 5 ) .
Expand and rearrange to get the quadratic equation 2 x 2 − 5 x − 3 = 0 .
Factor the quadratic equation as ( 2 x + 1 ) ( x − 3 ) = 0 .
Solve for x to find the solutions x = − 0.5 and x = 3 . The final answer is x = − 0.5 and x = 3 .
Explanation
Understanding the Problem We are given the equation 2 x − 5 3 = x . We need to find the solutions to this equation. The possible solutions are given as multiple choices.
Eliminating the Fraction First, we need to get rid of the fraction by multiplying both sides of the equation by ( 2 x − 5 ) . This gives us: 3 = x ( 2 x − 5 )
Expanding the Equation Next, we expand the right side of the equation: 3 = 2 x 2 − 5 x
Rearranging into Quadratic Form Now, we rearrange the equation into a standard quadratic form: 2 x 2 − 5 x − 3 = 0
Factoring the Quadratic We can solve this quadratic equation by factoring. We are looking for two numbers that multiply to 2 × − 3 = − 6 and add up to − 5 . These numbers are − 6 and 1 . So, we can rewrite the middle term as − 6 x + x : 2 x 2 − 6 x + x − 3 = 0
Factoring by Grouping Now, we factor by grouping: 2 x ( x − 3 ) + 1 ( x − 3 ) = 0 ( 2 x + 1 ) ( x − 3 ) = 0
Solving for x Setting each factor to zero, we get: 2 x + 1 = 0 or x − 3 = 0
Finding the Solutions Solving for x in each case: 2 x = − 1 ⇒ x = − 2 1 = − 0.5 x = 3
Final Solutions Therefore, the solutions to the equation are x = − 0.5 and x = 3 .
Examples
Quadratic equations like this one appear in various real-world scenarios, such as calculating the trajectory of a projectile, determining the dimensions of a rectangular area with a given perimeter and area, or modeling growth and decay processes. For instance, if you're designing a rectangular garden with a fixed area and want to minimize the amount of fencing needed, you might end up solving a quadratic equation to find the optimal dimensions.
