Answer: Choice C
x = 3, solution is not extraneous
Step-by-step explanation:
To solve for x follow these steps in the exact order shown
- square both sides
- subtract 1 from both sides
- divide both sides by 8
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
