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BloomLocke367

  • one year ago

Solve algebraically and confirm graphically. \((x+11)^2=121\)

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  1. BloomLocke367
    • one year ago
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    I got it down to \(x^2+22x=0\)

  2. BloomLocke367
    • one year ago
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    and I'm pretty sure I need to complete the square, right?

  3. jdoe0001
    • one year ago
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    I gather so....yes

  4. jdoe0001
    • one year ago
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    well... do you how to complete the square? :)

  5. BloomLocke367
    • one year ago
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    I did... lemme see if I can remember XD It's been almost an entire year since I've done it xp

  6. BloomLocke367
    • one year ago
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    and nope. I can't remember. Just tell me what to do first and I'm pretty sure I'll catch on quickly since I'm already somewhat familiar with it.

  7. mathstudent55
    • one year ago
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    You don't need to complete the square. Just factor out x on the left side.

  8. jdoe0001
    • one year ago
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    hmmm actually. right.. as @mathstudent55 said... you could just do common factoring

  9. BloomLocke367
    • one year ago
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    \(x^2+22x+121\). That's it expanded

  10. BloomLocke367
    • one year ago
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    but it is set equal to 121

  11. mathstudent55
    • one year ago
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    \((x+11)^2=121\) \(x^2 + 22x + 121 = 121\) \(x^2 + 22x = 0\) \(x(x + 22) = 0\) Now continue.

  12. geerky42
    • one year ago
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    Factor x out, then use the fact that if ab=0, then a=0 OR b=0.

  13. BloomLocke367
    • one year ago
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    ohhhhhhhhhhhhhhhhh

  14. jdoe0001
    • one year ago
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    \(x^2+22x=0\implies x(x+22)=0\implies \begin{cases} x=0\\ x+22=0 \end{cases}\)

  15. BloomLocke367
    • one year ago
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    so x=0 and x=-22

  16. mathstudent55
    • one year ago
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    Correct.

  17. BloomLocke367
    • one year ago
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    Thank you for making me remember stuff that I forgot XD

  18. BloomLocke367
    • one year ago
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    But now I don't know who to medal >.<

  19. jdoe0001
    • one year ago
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    well.. @BloomLocke367 that's non-important anyway :)

  20. mathstudent55
    • one year ago
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    The way the problem was given originally, it was already a completed square. You could have done this: \((x + 11)^2 = 121\) \(x + 11 = \pm \sqrt{121} \) \(x + 11 = \pm 11\) \(x + 11 = 11\) or \(x + 11 = -11\) \(x = 0\) or \(x = -11\)

  21. jim_thompson5910
    • one year ago
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    Why not just take the square root of both sides? \[\large (x+11)^2 = 121\] \[\large \sqrt{(x+11)^2} = \sqrt{121}\] \[\large |x+11| = 11 \ ...\ \text{Rule:} \sqrt{x^2} = |x| \text{ where x is a real number}\] \[\large x+11 = 11 \ \text{ or } \ x+11 = -11\] \[\large x= ??\ \text{ or } \ x = ??\]

  22. mathstudent55
    • one year ago
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    Now you need to do it graphically.

  23. BloomLocke367
    • one year ago
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    Wow, thank you for making me see things in many different ways. That helps a ton. Thank you all! I would give all of you a medal if I could.

  24. BloomLocke367
    • one year ago
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    oh my goodness these are getting more and more complex as they go on.

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