\frac{2}{d-2}=\frac{3 d}{4 d+12}
Answer (1)
Cross-multiply the given equation to eliminate fractions.
Expand and rearrange the equation into a standard quadratic form: 3 d 2 − 14 d − 24 = 0 .
Apply the quadratic formula to find the possible values of d .
Check for extraneous solutions by ensuring the denominators in the original equation are not zero. The solutions are 6 , − 3 4 .
Explanation
Problem Analysis We are given the equation d − 2 2 = 4 d + 12 3 d and we need to solve for d .
Cross-Multiplication and Expansion First, we cross-multiply to eliminate the fractions: 2 ( 4 d + 12 ) = 3 d ( d − 2 ) Expanding both sides, we get: 8 d + 24 = 3 d 2 − 6 d
Rearranging into Quadratic Form Now, we rearrange the equation into a quadratic equation: 3 d 2 − 6 d − 8 d − 24 = 0 3 d 2 − 14 d − 24 = 0
Applying the Quadratic Formula We solve the quadratic equation for d using the quadratic formula: d = 2 a − b ± b 2 − 4 a c where a = 3 , b = − 14 , and c = − 24 . Substituting these values, we get: d = 2 ( 3 ) 14 ± ( − 14 ) 2 − 4 ( 3 ) ( − 24 ) d = 6 14 ± 196 + 288 d = 6 14 ± 484 d = 6 14 ± 22
Finding the Solutions So, the two possible values for d are: d 1 = 6 14 + 22 = 6 36 = 6 d 2 = 6 14 − 22 = 6 − 8 = − 3 4 Thus, d = 6 or d = − 3 4 .
Checking for Extraneous Solutions We need to check if any of the solutions make the denominator of the original equation equal to zero. The denominators are d − 2 and 4 d + 12 = 4 ( d + 3 ) . Thus, d cannot be 2 or − 3 .
Since neither of our solutions, 6 and − 3 4 , are equal to 2 or − 3 , both are valid solutions.
Final Answer Therefore, the solutions are d = 6 and d = − 3 4 .
Examples
Imagine you are designing a bridge and need to calculate the load distribution on different supports. Solving rational equations like this one can help you determine the optimal placement and strength of the supports to ensure the bridge's stability. Similarly, in electrical circuit design, these equations can help calculate the resistance needed in different parts of the circuit to achieve a desired voltage or current flow. Understanding and solving such equations is crucial for ensuring the safety and efficiency of these real-world applications.
