x = 3, solution is not extraneous
Step-by-step explanation:
To solve for x follow these steps in the exact order shown
Notice how I'm following PEMDAS in reverse to undo each operation happening to x, so we can isolate it.
Here's what those steps look like:
[tex]\sqrt{8x+1} = 5\\\\(\sqrt{8x+1})^2 = 5^2\\\\8x+1 = 25\\\\8x = 25-1\\\\8x = 24\\\\x = 24/8\\\\x = 3\\\\[/tex]
We determine x = 3 is a solution. Whether it's extraneous or not depends on the next section.
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Someone claims "x = 3 is a solution to the given equation". To verify this claim, we replace each copy of x with 3. Then simplify. If x = 3 was indeed a solution, then both sides should boil down to the same number.
Here's what the scratch work looks like when verifying x = 3.
[tex]\sqrt{8x+1} = 5\\\\\sqrt{8*3+1} = 5\\\\\sqrt{24+1} = 5\\\\\sqrt{25} = 5\\\\5 = 5\\\\[/tex]
We get the same number on both sides. This confirms that x = 3 is indeed the solution. It's not extraneous because this value of x makes the original equation true. It shows why choice C is the final answer.
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Once we determine x = 3 is a solution, we don't need to check any other values.
However, I'll show what it looks like when x = 1/4 = 0.25 is plugged in to show what a non-solution looks like.
[tex]\sqrt{8x+1} = 5\\\\\sqrt{8*0.25+1} = 5\\\\\sqrt{2+1} = 5\\\\\sqrt{3} = 5\\\\1.73205 = 5\\\\[/tex]
Use a calculator for the last step to evaluate [tex]\sqrt{3}[/tex]
The two sides 1.73205 and 5 do not match up, so x = 1/4 is not a solution.
Another way to see this is to square both sides and [tex]\sqrt{3} = 5 \rightarrow (\sqrt{3})^2 = 5^2 \rightarrow 3 = 25[/tex] which is false.
So this is another way to see why x = 1/4 doesn't work.
Instead only x = 3 is the solution.
Once again, the final answer is choice C
