SOLUTION: 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

Algebra ->  Radicals -> SOLUTION: 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      Log On


   

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
then i added 9+4 and got >> 3x+1=13-6 sqrt(2x+4) +2x
then i subtrcated 2x from each side and got>>> x+1=13-6 sqrt(2x+4)
then i subtracted 1 from each side and got>> x=12-6sqrt(2x+4)
added 6sqrt(2x+4) to each side and got>>> 6sqrt(2x+4)+x=12
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
-72x-144 -144-72x
0=-96x+x^2
0=x^2+96
i wanted to know from her do i plug the number into
x=b +or- sqrt b^2-4(a)(c)divided by 2(a)
and i wanted to kno if u get a negative answer from this equation
x=b +or- sqrt b^2-4(a)(c)divided by 2(a) how do you check it because you cant square a negative in the calculator
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
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
then i added 9+4 and got >> 3x+1=13-6 sqrt(2x+4) +2x
then i subtrcated 2x from each side and got>>> x+1=13-6 sqrt(2x+4)
then i subtracted 1 from each side and got>> x=12-6sqrt(2x+4)
added 6sqrt(2x+4) to each side and got>>> 6sqrt(2x+4)+x=12
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
-72x-144 -144-72x
0=-96x+x^2
0=x^2+96 -96x not +
---------
You left off the x in 96x
x^2 - 96x = 0
x = 0
x = 96
Close, one slipup

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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
---
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
----
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.