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Mathematics 22 Online
OpenStudy (anonymous):

Select whether the equation has a solution or not. Explain every step pls

OpenStudy (anonymous):

\[\sqrt{x}+1=7-2\sqrt{x}\]

OpenStudy (usukidoll):

let's try find our x...

OpenStudy (usukidoll):

so there's a variable on the right. add to both sides to shift it to the left and then subtract 1 on both sides.

OpenStudy (usukidoll):

@MrNood is back! :D

OpenStudy (anonymous):

ok so now we have\[\sqrt{x}=6-2\sqrt{x}\]

OpenStudy (usukidoll):

ok good so far... what next? we want all variables to the left...

OpenStudy (anonymous):

\[3\sqrt{x}=6\]

OpenStudy (usukidoll):

yay... now divide both sides

OpenStudy (usukidoll):

the number parts

OpenStudy (anonymous):

Oh ok that was actually pretty easy x=4 I just messed up really early and it got complicated from there. Thanks

OpenStudy (anonymous):

So yeah it has a solution

OpenStudy (usukidoll):

you can plug 4 into the original equation to see if it equals.. I'm sure it does XD

OpenStudy (usukidoll):

when you plugged x = 4 back into the original equation, did you get 3?

OpenStudy (anonymous):

I got 2=2

OpenStudy (usukidoll):

how though? \[\sqrt{4}+1=7-2\sqrt{4}\] \[2+1=7-2(2)\] \[3=7-4\] \[3=3\]

OpenStudy (anonymous):

Oh well I didn't use the original equation I started when we had already subtracted 1 from each side but I see how u did it the original way too

OpenStudy (usukidoll):

to check to see if the solution works.. plug it back into the original equation... but I guess plugging it back earlier would work too... but it's best to do this to the original equation

OpenStudy (anonymous):

ok thanks

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