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anonymous
 5 years ago
I need help! the square root of 3x +1  the square root x1=2
anonymous
 5 years ago
I need help! the square root of 3x +1  the square root x1=2

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it would help out alot in understanding your question if you would use paratnthesis to indicate what would be "under" the radical. like this: sqrt(3x1)  sqrt(x) 1 = 0 otherwise, we dont know what your problem really is......

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(3x+1)sqrt(x1)=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0solutions are 1,5 dont know how to get there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{3x+1}\sqrt{x1}=2\quad /()^2\] \[3x+1+(x1)2\sqrt{(3x+1)(x1)}=4\] \[4x2\sqrt{(3x+1)(x1)}=4\qquad /4x\] \[2\sqrt{(3x+1)(x1)}=44x\qquad /:(2)\] \[\sqrt{(3x+1)(x1)}=2+2x\qquad /()^2\] \[(3x+1)(x1)=(2+2x)^2\] \[3x^22x1=4+4x^28x\qquad /3x^2+2x+1\] \[0=x^26x+5\] \[0=(x1)(x5)\] Solutions: \[x_1=1,\quad x_2=5\quad\checkmark\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0isolate a sqrt :) sqrt(x1) = 2 sqrt(3x+1) now square both side

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and remember to watch your signs.... i appear to have dropped one on accident

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you I am trying to comprehend all of this
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