Which of the following is an extraneous solution of $(45-3 x)^{\frac{1}{2}}=x-9 ?
A. $x=-12$
B. $x=-3$
C. $x=3$
D. $x=12$
Answer (2)
The extraneous solution for the equation ( 45 − 3 x ) 2 1 = x − 9 is x = 3 . This solution does not satisfy the original equation, thus it is considered extraneous. The valid solution is x = 12 .
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Square both sides of the equation ( 45 − 3 x ) 2 1 = x − 9 to get 45 − 3 x = ( x − 9 ) 2 .
Expand and rearrange to form the quadratic equation x 2 − 15 x + 36 = 0 .
Factor the quadratic equation to find the possible solutions x = 3 and x = 12 .
Check each solution in the original equation; x = 3 does not satisfy the original equation, while x = 12 does. Therefore, the extraneous solution is 3 .
Explanation
Understanding the Problem We are given the equation ( 45 − 3 x ) 2 1 = x − 9 and asked to find the extraneous solution from the options x = − 12 , x = − 3 , x = 3 , and x = 12 . An extraneous solution is a solution that arises during the solving process but does not satisfy the original equation.
Squaring Both Sides First, let's square both sides of the equation to eliminate the square root: ( 45 − 3 x ) 2 1 = x − 9 ( 45 − 3 x ) = ( x − 9 ) 2
Expanding the Equation Next, expand the right side of the equation: 45 − 3 x = x 2 − 18 x + 81
Rearranging to Quadratic Form Now, rearrange the equation to form a quadratic equation: x 2 − 18 x + 81 + 3 x − 45 = 0 x 2 − 15 x + 36 = 0
Solving the Quadratic Equation We can solve this quadratic equation by factoring: ( x − 3 ) ( x − 12 ) = 0 This gives us two possible solutions: x = 3 and x = 12 .
Checking x=3 Now, we need to check each possible solution in the original equation to see if it is an extraneous solution. Let's start with x = 3 : ( 45 − 3 ( 3 ) ) 2 1 = 3 − 9 ( 45 − 9 ) 2 1 = − 6 ( 36 ) 2 1 = − 6 6 = − 6 This is not true, so x = 3 is an extraneous solution.
Checking x=12 Now, let's check x = 12 : ( 45 − 3 ( 12 ) ) 2 1 = 12 − 9 ( 45 − 36 ) 2 1 = 3 ( 9 ) 2 1 = 3 3 = 3 This is true, so x = 12 is a valid solution.
Identifying Extraneous Solution Therefore, the extraneous solution is x = 3 .
Examples
When solving equations involving radicals, it's crucial to check for extraneous solutions. These solutions can arise because squaring both sides of an equation can introduce solutions that don't satisfy the original equation. For instance, consider a scenario where you're modeling the distance a projectile travels. If your equation involves a square root and you find two possible solutions, one positive and one negative, the negative solution might be extraneous since distance cannot be negative in the real world. Always verify your solutions in the original context to ensure they make sense.
