A student stated that there are two solutions to the equation [tex]$\sqrt{ }(3 x+7)=x-1$[/tex]
They are [tex]$x=-1$[/tex] and [tex]$x=6$[/tex].
Do you agree with the student? Explain why or why not.
A. I do not agree with the student because there is no real solution.
B. Checking the solutions in the original equation verifies [tex]$x=6$[/tex]; however, [tex]$x=-1$[/tex] is extraneous.
C. Checking the solutions in the original equation shows that 6 is an extraneous solution.
D. I agree with the student because both [tex]$x=-1$[/tex] and [tex]$x=6$[/tex] satisfy the original equation.
Answer (2)
The student is incorrect; only x = 6 is a valid solution to the equation 3 x + 7 = x − 1 . The solution x = − 1 is an extraneous solution and does not satisfy the original equation. Therefore, the correct answer is option B, as it verifies that x = 6 is valid while x = − 1 is not.
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Square both sides of the equation 3 x + 7 = x − 1 to get 3 x + 7 = ( x − 1 ) 2 .
Simplify and rearrange to form the quadratic equation x 2 − 5 x − 6 = 0 .
Factor the quadratic equation to find potential solutions x = 6 and x = − 1 .
Check both solutions in the original equation; x = 6 is a valid solution, while x = − 1 is extraneous. Therefore, only x = 6 is a solution. x = 6
Explanation
Understanding the Problem We are given the equation 3 x + 7 = x − 1 and a student claims that x = − 1 and x = 6 are the solutions. We need to verify this claim.
Squaring Both Sides To solve the equation, we first square both sides to eliminate the square root: ( 3 x + 7 ) 2 = ( x − 1 ) 2 This simplifies to: 3 x + 7 = x 2 − 2 x + 1
Rearranging the Equation Now, we rearrange the equation into a quadratic equation: x 2 − 2 x + 1 − 3 x − 7 = 0 x 2 − 5 x − 6 = 0
Solving the Quadratic Equation We can solve the quadratic equation by factoring: ( x − 6 ) ( x + 1 ) = 0 This gives us two possible solutions: x = 6 and x = − 1 .
Checking x=6 Now, we need to check if these solutions satisfy the original equation. Let's start with x = 6 : 3 ( 6 ) + 7 = 6 − 1 18 + 7 = 5 25 = 5 5 = 5 So, x = 6 is a valid solution.
Checking x=-1 Next, let's check x = − 1 : 3 ( − 1 ) + 7 = − 1 − 1 − 3 + 7 = − 2 4 = − 2 2 = − 2 This is not true, so x = − 1 is an extraneous solution.
Conclusion Therefore, the student is incorrect. Only x = 6 is a solution to the original equation. x = − 1 is an extraneous solution that arises from squaring both sides of the equation.
Examples
When solving equations involving square roots, it's crucial to check your solutions in the original equation. Squaring both sides can introduce extraneous solutions that don't actually satisfy the original equation. This concept is applicable in various fields, such as physics when dealing with energy equations or engineering when designing structures, where ensuring the validity of solutions is essential for accurate and reliable results.
