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anonymous

  • one year ago

I need help with algebra again

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  1. anonymous
    • one year ago
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    \[-\sqrt{6x+1}-5=-7-\sqrt{3x+1}\]

  2. UsukiDoll
    • one year ago
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    thinking

  3. anonymous
    • one year ago
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    lol ok

  4. UsukiDoll
    • one year ago
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    ok get the numbers to the right and those square roots to the left

  5. Michele_Laino
    • one year ago
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    hint: we can rewrite your equation as below: \[\Large 7 - 5 = \sqrt {6x + 1} - \sqrt {3x + 1} \]

  6. UsukiDoll
    • one year ago
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    half of me is like square both sides to get rid of those square roots. after that step

  7. Michele_Laino
    • one year ago
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    after a first square, we get: \[\Large 4 = 6x + 1 + 3x + 1 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \] please simplify @MathHater82

  8. UsukiDoll
    • one year ago
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    oh golly .

  9. anonymous
    • one year ago
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    o.o im lost

  10. Michele_Laino
    • one year ago
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    hint: here are the next steps: \[\Large \begin{gathered} 4 = 9x + 2 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\ \hfill \\ 2 = 9x - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\ \end{gathered} \]

  11. anonymous
    • one year ago
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    ok but how did you do the step before that

  12. UsukiDoll
    • one year ago
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    \[\Large 2= \sqrt {6x + 1} - \sqrt {3x + 1} \] \[\Large 2^2= (\sqrt {6x + 1} - \sqrt {3x + 1})^2 \] \[\Large 4= (\sqrt {6x + 1} - \sqrt {3x + 1})(\sqrt {6x + 1} - \sqrt {3x + 1}) \] and then expand ?

  13. UsukiDoll
    • one year ago
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    with foil?

  14. Michele_Laino
    • one year ago
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    since I have computed the square of both sides, like this: \[\Large {\left( {7 - 5} \right)^2} = {\left( {\sqrt {6x + 1} - \sqrt {3x + 1} } \right)^2}\]

  15. anonymous
    • one year ago
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    oh ok

  16. UsukiDoll
    • one year ago
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    good luck to that on the right hand side that's killer

  17. Michele_Laino
    • one year ago
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    next, we can write: \[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]

  18. Michele_Laino
    • one year ago
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    and, taking the square of both sides again, we have: \[\Large {\left( {2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} } \right)^2} = {\left( {9x - 2} \right)^2}\]

  19. Michele_Laino
    • one year ago
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    or, after a simplification: \[\Large 4\left( {18{x^2} + 9x + 1} \right) = 81{x^2} + 4 - 36x\]

  20. Michele_Laino
    • one year ago
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    I have applied the foil method inside the radical at left side

  21. Michele_Laino
    • one year ago
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    so, please continue @MathHater82

  22. UsukiDoll
    • one year ago
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    oh yes it's getting to be a bit easier. . . . after those square roots are gone.

  23. anonymous
    • one year ago
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    ok im just trying to follow this all on my white board i have here

  24. anonymous
    • one year ago
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    give me a sec

  25. anonymous
    • one year ago
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    \[36x+4=9x^2-36x+4\] \[72x=9x^2\]

  26. Michele_Laino
    • one year ago
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    ok! correct! now try to solve that equation

  27. anonymous
    • one year ago
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    x=8

  28. Michele_Laino
    • one year ago
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    hint: we have the subsequent steps: \[\Large \begin{gathered} 9{x^2} - 72x = 0 \hfill \\ \hfill \\ 9x\left( {x - 8} \right) = 0 \hfill \\ \end{gathered} \]

  29. UsukiDoll
    • one year ago
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    maybe we can shift the 9x^2 and se the equation equal to 0. whoops backwards 0 = 9x^2-72x

  30. UsukiDoll
    • one year ago
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    yes solve for x after what @Michele_Laino typed there's 2 x's to solve.. so first we have 9x=0 and then x-8=0

  31. Michele_Laino
    • one year ago
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    x=8 is correct! nevertheless also x=0 is a solution that's right! @UsukiDoll

  32. UsukiDoll
    • one year ago
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    x = 0,8 but uh oh only one x works

  33. Michele_Laino
    • one year ago
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    so, x=0 is a solution or not? @MathHater82

  34. Michele_Laino
    • one year ago
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    please recall this step: \[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]

  35. UsukiDoll
    • one year ago
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    I think plugging in x = 0 in the equation that we had started with is easier.

  36. Michele_Laino
    • one year ago
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    noew, the left side is positive, since it is the positive square root, so also the right side has to be positive. Namely we have: \[\Large 9x - 2 > 0\]

  37. Michele_Laino
    • one year ago
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    now*

  38. Michele_Laino
    • one year ago
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    or: \[\Large x > \frac{9}{2}\] now x=0, doesn't check that condition, so x=0 can not be an acceptable solution

  39. Michele_Laino
    • one year ago
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    |dw:1437567359392:dw|

  40. UsukiDoll
    • one year ago
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    also when x = 0 \[-\sqrt{1}-5=-7-\sqrt{1}\] -1-5=-7-1 \[-6 \neq -8\]

  41. Michele_Laino
    • one year ago
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    yes! that is another possible way to solve the question

  42. Michele_Laino
    • one year ago
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    mathematicians call x=0 an extraneous solution

  43. anonymous
    • one year ago
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    I don't really understand why x=0

  44. Michele_Laino
    • one year ago
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    since x=0 and x=8 are the solutions of this equation: \[\Large 9{x^2} - 72x = 0\]

  45. UsukiDoll
    • one year ago
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    we obtained two solutions, but 1 works.

  46. anonymous
    • one year ago
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    how did we obtain the solution 0 and which one worked

  47. UsukiDoll
    • one year ago
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    the equation \[\Large 9{x^2} - 72x = 0 \] got factored.. we can factor a 9x out

  48. UsukiDoll
    • one year ago
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    \[\large 9x(x-8)=0\] then we solve for each case 9x=0 and x-8=0 that leads us to x =0,8 but only 1 works.

  49. anonymous
    • one year ago
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    ok

  50. Michele_Laino
    • one year ago
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    furthermore. looking at your original equation also these conditions have to be checked: \[\Large \begin{gathered} 6x + 1 \geqslant 0 \hfill \\ 3x + 1 \geqslant 0 \hfill \\ \end{gathered} \] for existence of both radicals, which are automatically satisfied since: \[\Large 8 > \frac{9}{2}\]

  51. anonymous
    • one year ago
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    ok cool i get it thanks for the help

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