Which of the following is an extraneous solution of $\sqrt{4 x+41}=x+5$?
A. $x=-8$
B. $x=-2$
C. $x=2$
D. $x=8$
Answer (2)
The extraneous solution of the equation 4 x + 41 = x + 5 is x = − 8 . This value does not satisfy the original equation, while x = 2 does. Hence, the answer is − 8 .
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Square both sides of the equation 4 x + 41 = x + 5 to get 4 x + 41 = ( x + 5 ) 2 .
Simplify and rearrange to form the quadratic equation x 2 + 6 x − 16 = 0 .
Factor the quadratic equation as ( x + 8 ) ( x − 2 ) = 0 , yielding possible solutions x = − 8 and x = 2 .
Check the solutions in the original equation; x = − 8 is extraneous because it leads to 3 = − 3 , while x = 2 is a valid solution. Therefore, the extraneous solution is x = − 8 .
Explanation
Understanding the Problem We are given the equation 4 x + 41 = x + 5 and asked to find the extraneous solution. An extraneous solution is a solution that we find when solving the equation, but it does not satisfy the original equation.
Squaring Both Sides To solve the equation, we first square both sides to eliminate the square root: ( 4 x + 41 ) 2 = ( x + 5 ) 2
Simplifying the Equation Simplifying, we get: 4 x + 41 = x 2 + 10 x + 25
Forming a Quadratic Equation Rearranging the terms to form a quadratic equation, we have: x 2 + 10 x − 4 x + 25 − 41 = 0 x 2 + 6 x − 16 = 0
Factoring the Quadratic Equation Now, we factor the quadratic equation:We are looking for two numbers that multiply to -16 and add to 6. These numbers are 8 and -2. So, we can factor the equation as: ( x + 8 ) ( x − 2 ) = 0
Finding Possible Solutions Setting each factor equal to zero, we find the possible solutions: x + 8 = 0 ⇒ x = − 8 x − 2 = 0 ⇒ x = 2
Checking for Extraneous Solutions Now, we need to check if these solutions are extraneous by substituting them back into the original equation.First, let's check x = − 8 : 4 ( − 8 ) + 41 = − 8 + 5 − 32 + 41 = − 3 9 = − 3 3 = − 3 This is false, so x = − 8 is an extraneous solution.Next, let's check x = 2 : 4 ( 2 ) + 41 = 2 + 5 8 + 41 = 7 49 = 7 7 = 7 This is true, so x = 2 is a valid solution.
Conclusion Therefore, the extraneous solution is x = − 8 .
Examples
Radical equations appear in various fields, such as physics and engineering, when dealing with relationships involving square roots or other radicals. For example, when calculating the velocity of an object in free fall or determining the period of a pendulum, radical equations may arise. Understanding how to solve these equations and identify extraneous solutions is crucial for obtaining accurate and meaningful results in these applications. Extraneous solutions can lead to incorrect predictions or designs, highlighting the importance of verifying solutions in the original equation.
