Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
[tex]\frac{5}{x}-\frac{15}{x-2}+4=0[/tex]
Answer (2)
Multiply both sides by x ( x − 2 ) to eliminate fractions.
Simplify the equation to the quadratic form: 2 x 2 − 9 x − 5 = 0 .
Use the quadratic formula to find the solutions: x = 4 9 ± 11 .
The real solutions are x = 5 and x = − 2 1 , so the final answer is − 2 1 , 5 .
Explanation
Understanding the Problem We are given the equation x 5 − x − 2 15 + 4 = 0 . Our goal is to find all real solutions for x . We need to be mindful that x cannot be 0 or 2, as these values would make the denominators of the fractions equal to zero, which is not allowed.
Eliminating Fractions To solve the equation, we first eliminate the fractions by multiplying both sides of the equation by x ( x − 2 ) . This gives us: 5 ( x − 2 ) − 15 x + 4 x ( x − 2 ) = 0
Expanding and Simplifying Next, we expand and simplify the equation: 5 x − 10 − 15 x + 4 x 2 − 8 x = 0 4 x 2 − 18 x − 10 = 0
Further Simplification We can divide the entire equation by 2 to simplify it further: 2 x 2 − 9 x − 5 = 0
Solving the Quadratic Equation Now, we solve the quadratic equation 2 x 2 − 9 x − 5 = 0 for x . We can use the quadratic formula: x = 2 a − b ± b 2 − 4 a c where a = 2 , b = − 9 , and c = − 5 . Plugging in these values, we get: x = 2 ( 2 ) 9 ± ( − 9 ) 2 − 4 ( 2 ) ( − 5 ) x = 4 9 ± 81 + 40 x = 4 9 ± 121 x = 4 9 ± 11
Finding the Solutions This gives us two possible solutions: x 1 = 4 9 + 11 = 4 20 = 5 x 2 = 4 9 − 11 = 4 − 2 = − 2 1 So, the solutions are x = 5 and x = − 2 1 .
Checking for Extraneous Solutions We need to check if these solutions are valid by making sure they don't make the original denominators zero. Since x = 5 and x = − 2 1 are not equal to 0 or 2, both solutions are valid.
Final Answer Therefore, the real solutions of the equation are 5 and − 2 1 .
Examples
Imagine you are designing a bridge and need to calculate the stress distribution on its supports. Equations similar to the one we solved can arise when modeling the forces and ensuring the bridge's stability. Finding the real solutions helps engineers determine critical points where the stress is at its maximum or minimum, ensuring the structure's safety and longevity. This type of problem also appears in circuit analysis when determining the values of resistors or capacitors needed to achieve a specific circuit behavior. By solving such equations, engineers can design circuits that meet desired specifications.
The real solutions of the equation x 5 − x − 2 15 + 4 = 0 are 5 and − 2 1 . Both values are checked and do not cause any division by zero. Therefore, the final answer is 5 , − 2 1 .
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