Answered

Given the equation [tex]\sqrt{8x+1}=5[/tex], solve for [tex]x[/tex] and identify if it is an extraneous solution.

A. [tex]x=\frac{1}{4}[/tex], solution is not extraneous
B. [tex]x=\frac{1}{4}[/tex], solution is extraneous
C. [tex]x=3[/tex], solution is not extraneous
D. [tex]x=3[/tex], solution is extraneous



Answer :

GinnyAnswer
To solve the equation [tex]\(\sqrt{8x + 1} = 5\)[/tex] for [tex]\(x\)[/tex], follow these steps:

1. Isolate the square root:
[tex]\[\sqrt{8x + 1} = 5\][/tex]

2. Square both sides to eliminate the square root:
[tex]\[(\sqrt{8x + 1})^2 = 5^2\][/tex]
This simplifies to:
[tex]\[8x + 1 = 25\][/tex]

3. Solve the resulting linear equation:
Subtract 1 from both sides:
[tex]\[8x = 24\][/tex]
Then divide by 8:
[tex]\[x = 3\][/tex]

4. Check for extraneous solutions:
Substitute [tex]\(x = 3\)[/tex] back into the original equation to ensure it holds true.
[tex]\[\sqrt{8(3) + 1} = 5\][/tex]
Simplify inside the square root:
[tex]\[\sqrt{24 + 1} = 5\][/tex]
[tex]\[\sqrt{25} = 5\][/tex]
Since this is true ([tex]\(5 = 5\)[/tex]), [tex]\(x = 3\)[/tex] is a valid solution and is not extraneous.

To summarize, [tex]\(x = 3\)[/tex] is the solution, and it is not an extraneous solution.

So the correct option is:
[tex]\[ \boxed{x=3, \text{solution is not extraneous}} \][/tex]

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