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anonymous

  • one year ago

Can someone just help me solve this please?? i need an extraneous solution and it solved. i chose this but im not sure its correct. 2√x+5+7=4

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  1. stefrheart
    • one year ago
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    Is that the original equation?

  2. anonymous
    • one year ago
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    no it said to choose your own using a√x+b+c=d

  3. stefrheart
    • one year ago
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    Ah I understand hold for a minute please

  4. anonymous
    • one year ago
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    okay.

  5. stefrheart
    • one year ago
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    I dont think it has an extraneous solution because I got x=16

  6. anonymous
    • one year ago
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    well could you help me find one with one please? using the a√x+b+c=d formula thingy

  7. stefrheart
    • one year ago
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    Sure!

  8. stefrheart
    • one year ago
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    So you do understand what an extraneous solution is correct?

  9. anonymous
    • one year ago
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    i think it is how it solves the problem but if you plug it back in it doesnt work out

  10. anonymous
    • one year ago
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    like if you solve the radical but if you plug it in to check the answer will not work out

  11. stefrheart
    • one year ago
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    Yes that is correct! :)

  12. anonymous
    • one year ago
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    im not sure how to like create and solve a radical like that though... ;(

  13. anonymous
    • one year ago
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    Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model. a√x+b+c=d Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

  14. stefrheart
    • one year ago
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    is it \[a \sqrt{x+b+c}=d\] or \[a \sqrt{x} + b + c=d\]

  15. phi
    • one year ago
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    try something like \[ \sqrt{x+2} + 1 = -1 \]

  16. anonymous
    • one year ago
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    the first one @stefrheart

  17. anonymous
    • one year ago
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    @phi the only problem is is that i don't know how to solve that equation to check if it has an extraneous solution

  18. stefrheart
    • one year ago
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    To solve that all you need to do is subtract the one on both sides to create \[\sqrt{x+2}=-2\] and then square both sides \[(\sqrt{x+2})^{2}=(-2)^{}\] and then solve

  19. stefrheart
    • one year ago
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    just for phi's equation

  20. dinamix
    • one year ago
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    x=-8 @stefrheart

  21. dinamix
    • one year ago
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    @kenzi041

  22. anonymous
    • one year ago
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    i dont even know. i never learned radicals

  23. stefrheart
    • one year ago
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    oh you dont know what a radical is or how to work with it?

  24. anonymous
    • one year ago
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    we had a large learning gap because they changed curriculums

  25. anonymous
    • one year ago
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    not really.

  26. stefrheart
    • one year ago
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    Ah okay so you should learn what a radical is first then, before proceeding with the problem

  27. anonymous
    • one year ago
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    oh.

  28. stefrheart
    • one year ago
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    It would make alot more sense before solving this. Do you know what a \[x ^{2}\] is? It would be easier to teach you if you knew what it is

  29. anonymous
    • one year ago
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    x squared?

  30. stefrheart
    • one year ago
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    yes you know what to do with it if someone gave you the equation \[4^{2}\] ?

  31. anonymous
    • one year ago
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    16, i know like how to do that i just dont understand how to use like larger equations like a√x+b+c=d

  32. stefrheart
    • one year ago
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    Ah okay so you know how to do that. Well the larger equations are just the same ting as a regular square root.

  33. anonymous
    • one year ago
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    okay, so how do i find one with an extraneous solution

  34. anonymous
    • one year ago
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    because i have one for without but i just need one with one

  35. stefrheart
    • one year ago
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    I think I may have one hold on

  36. anonymous
    • one year ago
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    ok

  37. dinamix
    • one year ago
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    x=-8 u have check in the equation ans see it

  38. stefrheart
    • one year ago
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    Whats the extraneous solution then? @dinamix

  39. anonymous
    • one year ago
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    like how do you plug it in to check if it has one @dinamix

  40. anonymous
    • one year ago
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    i already tried but it was in a different format than what the question asked

  41. stefrheart
    • one year ago
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    Yeah I noticed ​​

  42. anonymous
    • one year ago
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    and i got more confused..

  43. stefrheart
    • one year ago
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    Hmmm im trying to figure out equations because in order for it to have both an actual and extraneous there needs to be 2 solutions which is a \[x ^{2}+bx+c\] unless we can do that you cant get both an actual and extraneous

  44. anonymous
    • one year ago
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    no it needs to just be a extraneous solution.. you write two problems and one should be regular and one should have an extraneous solution

  45. stefrheart
    • one year ago
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    Ohhh then I already have one then

  46. stefrheart
    • one year ago
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    \[-5\sqrt{-3x-5}=-25\]

  47. anonymous
    • one year ago
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    but doesnt it have to be x+b+c=d

  48. stefrheart
    • one year ago
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    if you wwant it you can just put -3x+-5+0 for under the radical

  49. anonymous
    • one year ago
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    okay, one sec im checking a problem i just did.

  50. stefrheart
    • one year ago
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    I will be back in a few minutes

  51. anonymous
    • one year ago
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    i think i found one, the problem i chose originally

  52. anonymous
    • one year ago
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    and okay

  53. stefrheart
    • one year ago
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    AH okay

  54. stefrheart
    • one year ago
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    Message me if need you still need help

  55. anonymous
    • one year ago
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    thanks for helping

  56. anonymous
    • one year ago
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    how do i give u a medal?

  57. stefrheart
    • one year ago
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    @kenzi041 click the best response

  58. anonymous
    • one year ago
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    @stefrheart i failed, but thanks anyway

  59. stefrheart
    • one year ago
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    Aw im sorry :( @kenzi041

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