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)
Square both sides of the equation 4 x + 41 = x + 5 to get 4 x + 41 = ( x + 5 ) 2 .
Simplify and rearrange into a quadratic equation: x 2 + 6 x − 16 = 0 .
Solve the quadratic equation by factoring: ( x + 8 ) ( x − 2 ) = 0 , which gives x = − 8 or x = 2 .
Check for extraneous solutions by substituting back into the original equation. x = − 8 is extraneous, while x = 2 is a valid solution. The extraneous solution is − 8 .
Explanation
Understanding the Problem We are given the equation 4 x + 41 = x + 5 and asked to find the extraneous solution from the options x = − 8 , x = − 2 , x = 2 , and x = 8 . An extraneous solution is a solution that arises when solving an equation but does not satisfy the original equation.
Eliminating the Square Root To find the extraneous solution, we first need to solve the equation. We square both sides of the equation to eliminate the square root: ( 4 x + 41 ) 2 = ( x + 5 ) 2 Simplifying, we get: 4 x + 41 = x 2 + 10 x + 25
Rearranging into Quadratic Form Now, we rearrange the equation into a quadratic equation: x 2 + 10 x + 25 − 4 x − 41 = 0 x 2 + 6 x − 16 = 0
Solving the Quadratic Equation We can solve this quadratic equation by factoring. We look for two numbers that multiply to -16 and add to 6. These numbers are 8 and -2. So, we can factor the quadratic equation as: ( x + 8 ) ( x − 2 ) = 0 This gives us two possible solutions: x = − 8 or x = 2
Checking for Extraneous Solutions Now we need to check if these solutions are extraneous by substituting them back into the original equation 4 x + 41 = x + 5 . For x = − 8 : 4 ( − 8 ) + 41 = − 32 + 41 = 9 = 3 And ( − 8 ) + 5 = − 3 Since 3 = − 3 , x = − 8 is an extraneous solution. For x = 2 : 4 ( 2 ) + 41 = 8 + 41 = 49 = 7 And 2 + 5 = 7 Since 7 = 7 , x = 2 is a valid solution.
Identifying the Extraneous Solution Therefore, the extraneous solution is x = − 8 .
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 example, consider designing a bridge where the length of a support cable is modeled by a square root function. If you solve for the length and obtain two possible values, one might be extraneous (e.g., a negative length), which is physically impossible. Therefore, checking solutions against the original physical constraints is essential to ensure the design is valid and safe.
The extraneous solution to the equation 4 x + 41 = x + 5 is x = − 8 . We found the solutions by squaring the equation and checking each solution in the original equation. Only x = − 8 did not satisfy the original equation.
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