Question 216086
sorry but i'am not sure what this kind of problem is
what is x?
sqrt (3x+1)+ sqrt (2x+4)= 3 
i subtracted>>>sqrt (2x+4) from both sides and got >>>>sqrt (3x+1)= 3 -sqrt (2x+4)
then i squared both sides to remove the radical and got>> 
3x+1= (3-(sqrt(2x+4)(3-sqrt(2x+4)then i did that and got
3x+1= 9-6 sqrt(2x+4) +2x+4
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then i subtracted x from both sides and got>>> 6sqrt(2x+4)=12-x
then i squared both sides to remove the radical n got
36(2x+4)=(12-x)(12-x)
72x+144=144-12x-12x+x^2
72x+144= 144-24x+x^2
x^2-96x = 0
x(x-96) = 0
x = 0 or x = 96
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At this point you have to check these "solutions"
in the original equation.  If x = 0 gives you
a true statement x=0 IS a solution.
Similarly you need to check x = 96.
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Cheers,
Stan H.