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√8x + 1 =5 show work

Mathematics
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is the problem unclear or is the problem exactly as stated?
\[\sqrt{8}x+1=5 => \sqrt{8}x=5-1=> \sqrt{8}x=4 => (\sqrt{8}x)^2=4^2\] \[=>8x^2=16 => x^2=2 => x = \pm \sqrt{2}\] But only \[x=\sqrt{2} \text{ is a solution} \]

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√x - 7 + 5 = 11 show work
\[\text{for some reason I think you meant } \sqrt{8x+1}=5\] square both sides \[8x+1=5^2\] 8x+1=25 subtract 1 on both sides 8x=25-1 8x=24 divide by 8 on both sides x=3 check your result: \[\sqrt{8(3)+1}=\sqrt{25}=5\] both sides are the same x=3 is a solution!
Wouldn't x=2 because 2*8=16 sqt16=4 4+1=5

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