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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
Let's solve the equation [tex]\(\sqrt{8x + 1} = 5\)[/tex] step-by-step and identify if the solution is extraneous or not.

1. Isolate the square root term:
[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 for [tex]\(x\)[/tex]:
Subtract 1 from both sides:
[tex]\[ 8x = 24 \][/tex]

Divide both sides by 8:
[tex]\[ x = 3 \][/tex]

4. Check if the solution is extraneous by substituting [tex]\(x = 3\)[/tex] back into the original equation:
[tex]\[ \sqrt{8(3) + 1} = 5 \][/tex]
Simplify inside the square root:
[tex]\[ \sqrt{24 + 1} = 5 \][/tex]
[tex]\[ \sqrt{25} = 5 \][/tex]
Since [tex]\(\sqrt{25} = 5\)[/tex] is true, the solution [tex]\(x = 3\)[/tex] satisfies the original equation.

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

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

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