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



Answer :

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

### Step 1: Isolate the square root
The equation given is:
[tex]\[ \sqrt{8x + 1} = 5 \][/tex]

### Step 2: Square both sides
By squaring both sides of the equation, we eliminate the square root:
[tex]\[ (\sqrt{8x + 1})^2 = 5^2 \][/tex]
[tex]\[ 8x + 1 = 25 \][/tex]

### Step 3: Solve for [tex]\(x\)[/tex]
Next, subtract 1 from both sides to isolate the term with [tex]\(x\)[/tex]:
[tex]\[ 8x = 25 - 1 \][/tex]
[tex]\[ 8x = 24 \][/tex]

Then, divide both sides by 8:
[tex]\[ x = \frac{24}{8} \][/tex]
[tex]\[ x = 3 \][/tex]

### Step 4: Verify the solution
To check if [tex]\(x = 3\)[/tex] is an extraneous solution, substitute [tex]\(x = 3\)[/tex] back into the original equation:
[tex]\[ \sqrt{8(3) + 1} = \sqrt{24 + 1} = \sqrt{25} = 5 \][/tex]

Since the left-hand side equals the right-hand side (5 = 5), [tex]\(x = 3\)[/tex] is not an extraneous solution.

### Conclusion
The solution [tex]\(x = 3\)[/tex] satisfies the original equation and is not extraneous. Therefore, the correct option is:

[tex]\(x=3\)[/tex], solution is not extraneous.