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anonymous

  • one year ago

Please help with Algebra will award medal

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  1. anonymous
    • one year ago
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    Task 2 Part 1. 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. Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous. Part 3. Explain why the first equation has an extraneous solution and the second does not.

  2. anonymous
    • one year ago
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    An extraneous solution is a fake result. You can go through and solve the equation, but when you check the answer it won't work. The easiest thing to do for both equations is to make a = 1 so you don't have to worry about division. Then you're left with \(\sqrt{x+b}+c=d\) Subtract c from both sides to get \(\sqrt{x+b}=d-c\) For the extraneous solution equation, you need d-c to be negative. For the other one d-c will be positive.

  3. anonymous
    • one year ago
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    so I can choose any number or does It have to be specific

  4. anonymous
    • one year ago
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    You can choose any number

  5. anonymous
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    so I plug in the numbers ad solve

  7. anonymous
    • one year ago
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    I couldn't see anything in that file. Can you post a screenshot or use the draw tool?

  8. anonymous
    • one year ago
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    ok I plugged In numbers a radical sign 18 +24+32=46

  9. anonymous
    • one year ago
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    so I subtract the 32 from both sides now right

  10. anonymous
    • one year ago
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    you're still supposed to have a radical sign and an x. The only numbers you're replacing are a,b,c, and d. For example √(x-3)+4=20

  11. anonymous
    • one year ago
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    For that equation a =1, b=-3, c = 4, and d = 20. Check d - c: d - c = 20 - 4 = 16 This is positive so that's an equation without an extraneous solution

  12. anonymous
    • one year ago
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    So I don't need to add a number for x

  13. anonymous
    • one year ago
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    right. x stays

  14. anonymous
    • one year ago
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    So so I subtract 4 from both sides and then now I have 1\ x+3=16 sorry I don't have a radical sign on my keyboard lol

  15. anonymous
    • one year ago
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    So I keep simplifying the equation from there

  16. anonymous
    • one year ago
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    yes after you subtract 4 square both sides

  17. anonymous
    • one year ago
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    Ok now I have 1/ x + 3 =16

  18. anonymous
    • one year ago
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    Than I subtracted 3 from both sides and now it's just 1/x=13

  19. anonymous
    • one year ago
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    no, square both sides first then subtract 3. You can't subtract the 3 while it's under the radical.|dw:1437332991847:dw|

  20. anonymous
    • one year ago
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    Square the 3 also or no

  21. anonymous
    • one year ago
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    no, the whole thing under the radical is squared. The √ and ² cancel each other \[\sqrt{x-3}^2=x-3\]

  22. anonymous
    • one year ago
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    Oh ok so x + 3 =256 so do I subtract the three now or leave it as it is

  23. anonymous
    • one year ago
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    subtract the 3

  24. anonymous
    • one year ago
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    x=233

  25. anonymous
    • one year ago
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    253? Just realized we switched for x-3 to x+3 when we were solving it. Just change the sign in the original equation to +

  26. anonymous
    • one year ago
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    Ok lol yea it's 253

  27. anonymous
    • one year ago
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    For the extraneous equation, just swap c and d and go through the same steps √(x-3)+20=4

  28. anonymous
    • one year ago
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    Ok I just finished that part now how would I explain why the first one was extraneous do I just say because the answer doesn't go with the equation

  29. anonymous
    • one year ago
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    yes, the answer doesn't check

  30. anonymous
    • one year ago
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    Ok thank you

  31. anonymous
    • one year ago
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    you're welcome

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