Cross-multiply the equation to eliminate fractions: w 2 = 4 ( 2 w − 3 ) .
Expand and rearrange the equation into a standard quadratic form: w 2 − 8 w + 12 = 0 .
Factor the quadratic equation: ( w − 2 ) ( w − 6 ) = 0 .
Solve for w and check for validity: w = 2 and w = 6 .
The solutions are w = 2 b r a ce w = 6 .
Explanation
Problem Analysis We are given the equation 2 w − 3 w = w 4 and we need to find the solutions for w . First, we need to identify any values of w that would make the denominators zero, as these values would be excluded from our solutions.
Cross-Multiplication and Expansion To solve the equation, we can cross-multiply to eliminate the fractions. This gives us: w 2 = 4 ( 2 w − 3 ) Expanding the right side, we get: w 2 = 8 w − 12
Forming a Quadratic Equation Now, we rearrange the equation to form a quadratic equation by moving all terms to one side: w 2 − 8 w + 12 = 0
Factoring the Quadratic Equation Next, we factor the quadratic equation: ( w − 2 ) ( w − 6 ) = 0
Solving for w Setting each factor equal to zero, we find the possible solutions for w :
w − 2 = 0 ⇒ w = 2 w − 6 = 0 ⇒ w = 6
Checking for Validity Now, we need to check if these solutions are valid by ensuring they don't make the denominators in the original equation equal to zero. The denominators are 2 w − 3 and w .
If w = 2 , then 2 w − 3 = 2 ( 2 ) − 3 = 4 − 3 = 1 , which is not zero. Also, w = 2 is not zero. If w = 6 , then 2 w − 3 = 2 ( 6 ) − 3 = 12 − 3 = 9 , which is not zero. Also, w = 6 is not zero. Therefore, both w = 2 and w = 6 are valid solutions.
Final Answer The solutions to the equation 2 w − 3 w = w 4 are w = 2 and w = 6 .
Examples
Imagine you're designing a rectangular garden where the ratio of the width to a certain expression involving the width must equal a specific value. Solving such an equation helps you determine the exact dimensions that satisfy the design requirements, ensuring your garden looks exactly as planned. This type of problem arises in various fields, including engineering, architecture, and computer graphics, where proportional relationships must be maintained.
The solutions to the equation 2 w − 3 w = w 4 are w = 2 and w = 6 . Therefore, the correct choice is D: w = 2 and w = 6 .
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