Sure, let's solve the equation [tex]\(\sqrt{x + 9} - 4 = 1\)[/tex] step by step.
1. Isolate the square root:
[tex]\[\sqrt{x + 9} - 4 = 1\][/tex]
Add 4 to both sides of the equation to isolate the square root term:
[tex]\[\sqrt{x + 9} = 5\][/tex]
2. Square both sides:
To eliminate the square root, square both sides of the equation:
[tex]\[(\sqrt{x + 9})^2 = 5^2\][/tex]
This simplifies to:
[tex]\[x + 9 = 25\][/tex]
3. Solve for [tex]\(x\)[/tex]:
Subtract 9 from both sides to solve for [tex]\(x\)[/tex]:
[tex]\[x = 25 - 9\][/tex]
Therefore:
[tex]\[x = 16\][/tex]
Now, we need to verify whether [tex]\[x = 16\][/tex] is not an extraneous solution by substituting it back into the original equation.
4. Check for extraneous solutions:
Substitute [tex]\(x = 16\)[/tex] back into the original equation [tex]\(\sqrt{x + 9} - 4 = 1\)[/tex]:
[tex]\[\sqrt{16 + 9} - 4 = 1\][/tex]
Simplify the expression inside the square root:
[tex]\[\sqrt{25} - 4 = 1\][/tex]
Since [tex]\(\sqrt{25} = 5\)[/tex], we have:
[tex]\[5 - 4 = 1\][/tex]
Which simplifies to:
[tex]\[1 = 1\][/tex]
This is true, so [tex]\(x = 16\)[/tex] is a valid solution and not extraneous.
5. Consider any other possible solutions:
Since squaring an equation can sometimes introduce extraneous solutions, it's important to verify each solution this way. In this problem, the algebraic steps yield a single valid solution.
Therefore, the solution to the equation [tex]\(\sqrt{x+9} - 4 = 1\)[/tex] is [tex]\(x = 16\)[/tex], and it is not an extraneous solution.
To answer the specific options given:
- [tex]\(x = 34\)[/tex], solution is extraneous (Not applicable here)
- [tex]\(x = 16\)[/tex], solution is not extraneous (Correct)
So, the correct conclusion based on the verification is that [tex]\(x = 16\)[/tex] is not extraneous and is indeed the valid solution.
