The solution to the equation 1 − 3 x = x + 3 is x = − 1 . The other potential solution, x = − 8 , is extraneous and does not satisfy the original equation. Thus, the correct answer is C . x = − 1 .
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Square both sides of the equation: ( 1 − 3 x ) 2 = ( x + 3 ) 2 , which simplifies to 1 − 3 x = x 2 + 6 x + 9 .
Rearrange the equation into a quadratic equation: x 2 + 9 x + 8 = 0 .
Solve the quadratic equation by factoring: ( x + 1 ) ( x + 8 ) = 0 , yielding potential solutions x = − 1 and x = − 8 .
Check for extraneous solutions: x = − 1 is valid, while x = − 8 is extraneous. The final answer is − 1 .
Explanation
Understanding the Problem We are given the equation 1 − 3 x = x + 3 . Our goal is to find the solution(s) for x and check for extraneous solutions.
Eliminating the Square Root First, we square both sides of the equation to eliminate the square root: ( 1 − 3 x ) 2 = ( x + 3 ) 2 This simplifies to: 1 − 3 x = x 2 + 6 x + 9
Rearranging into Quadratic Form Next, we rearrange the equation into a quadratic equation: x 2 + 6 x + 9 = 1 − 3 x x 2 + 9 x + 8 = 0
Solving the Quadratic Equation Now, we solve the quadratic equation for x by factoring: x 2 + 9 x + 8 = ( x + 1 ) ( x + 8 ) = 0 This gives us two possible solutions: x = − 1 and x = − 8 .
Checking for Extraneous Solutions We must check both solutions in the original equation 1 − 3 x = x + 3 to eliminate extraneous solutions.
For x = − 1 : 1 − 3 ( − 1 ) = − 1 + 3 1 + 3 = 2 4 = 2 2 = 2 So x = − 1 is a valid solution.
For x = − 8 : 1 − 3 ( − 8 ) = − 8 + 3 1 + 24 = − 5 25 = − 5 5 = − 5 This is false, so x = − 8 is an extraneous solution.
Final Answer Therefore, the only solution to the equation 1 − 3 x = x + 3 is x = − 1 .
Examples
Consider a scenario where you are designing a suspension bridge. The equation 1 − 3 x = x + 3 can be analogous to a simplified model of the forces and tensions acting on the bridge. Solving such equations helps engineers determine critical parameters for the bridge's stability and safety. For instance, 'x' might represent a normalized measure of tension, and finding its valid solution ensures that the tension remains within acceptable limits to prevent structural failure. This ensures the bridge can withstand the loads it's designed to carry, demonstrating the practical importance of solving equations in engineering.