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√(8x+1)=5
i need to solve for x and identify if it is an extraneous solution.
Is it \(\huge \sqrt{8x+1}=5\) ?
yes
\(\huge (\sqrt{8x+1})^2=(5)^2\)
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where did you get the power of 2?
So you square both sides to remove the radical sign.
ohhhh nvm wait
i understand why lol
It would now be: \(\huge 8x+1=25\)
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so x equals 3??
Yes! To check if it's an extraneous solution, plug in \(\huge x=3\) to the equation.
\(\huge \sqrt{8(3)+1}=5; x=3 \)
whats the difference between non extraneous and extraneous?
extraneous: not a real solution non extraneous: real solution / the x/root is true
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\(\huge \sqrt{24+1}=5\) \(\huge \sqrt{25}=5\) Is this true?
yeah
ok thank you so much!!!
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