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anonymous

  • 5 years ago

I need help! the square root of 3x +1 - the square root x-1=2

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

  2. anonymous
    • 5 years ago
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    ok

  3. anonymous
    • 5 years ago
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    sqrt(3x+1)-sqrt(x-1)=2

  4. anonymous
    • 5 years ago
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    solutions are 1,5 dont know how to get there

  5. anonymous
    • 5 years ago
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    \[\sqrt{3x+1}-\sqrt{x-1}=2\quad /()^2\] \[3x+1+(x-1)-2\sqrt{(3x+1)(x-1)}=4\] \[4x-2\sqrt{(3x+1)(x-1)}=4\qquad /-4x\] \[-2\sqrt{(3x+1)(x-1)}=4-4x\qquad /:(-2)\] \[\sqrt{(3x+1)(x-1)}=-2+2x\qquad /()^2\] \[(3x+1)(x-1)=(-2+2x)^2\] \[3x^2-2x-1=4+4x^2-8x\qquad /-3x^2+2x+1\] \[0=x^2-6x+5\] \[0=(x-1)(x-5)\] Solutions: \[x_1=1,\quad x_2=5\quad\checkmark\]

  6. amistre64
    • 5 years ago
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    isolate a sqrt :) sqrt(x-1) = 2 -sqrt(3x+1) now square both side

  7. amistre64
    • 5 years ago
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    and remember to watch your signs.... i appear to have dropped one on accident

  8. anonymous
    • 5 years ago
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    thank you I am trying to comprehend all of this

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