BloomLocke367
  • BloomLocke367
Solve algebraically and confirm graphically. \((x+11)^2=121\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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BloomLocke367
  • BloomLocke367
I got it down to \(x^2+22x=0\)
BloomLocke367
  • BloomLocke367
and I'm pretty sure I need to complete the square, right?
jdoe0001
  • jdoe0001
I gather so....yes

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More answers

jdoe0001
  • jdoe0001
well... do you how to complete the square? :)
BloomLocke367
  • BloomLocke367
I did... lemme see if I can remember XD It's been almost an entire year since I've done it xp
BloomLocke367
  • BloomLocke367
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.
mathstudent55
  • mathstudent55
You don't need to complete the square. Just factor out x on the left side.
jdoe0001
  • jdoe0001
hmmm actually. right.. as @mathstudent55 said... you could just do common factoring
BloomLocke367
  • BloomLocke367
\(x^2+22x+121\). That's it expanded
BloomLocke367
  • BloomLocke367
but it is set equal to 121
mathstudent55
  • mathstudent55
\((x+11)^2=121\) \(x^2 + 22x + 121 = 121\) \(x^2 + 22x = 0\) \(x(x + 22) = 0\) Now continue.
geerky42
  • geerky42
Factor x out, then use the fact that if ab=0, then a=0 OR b=0.
BloomLocke367
  • BloomLocke367
ohhhhhhhhhhhhhhhhh
jdoe0001
  • jdoe0001
\(x^2+22x=0\implies x(x+22)=0\implies \begin{cases} x=0\\ x+22=0 \end{cases}\)
BloomLocke367
  • BloomLocke367
so x=0 and x=-22
mathstudent55
  • mathstudent55
Correct.
BloomLocke367
  • BloomLocke367
Thank you for making me remember stuff that I forgot XD
BloomLocke367
  • BloomLocke367
But now I don't know who to medal >.<
jdoe0001
  • jdoe0001
well.. @BloomLocke367 that's non-important anyway :)
mathstudent55
  • mathstudent55
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 \(x = -11\)
jim_thompson5910
  • jim_thompson5910
Why not just take the square root of both sides? \[\large (x+11)^2 = 121\] \[\large \sqrt{(x+11)^2} = \sqrt{121}\] \[\large |x+11| = 11 \ ...\ \text{Rule:} \sqrt{x^2} = |x| \text{ where x is a real number}\] \[\large x+11 = 11 \ \text{ or } \ x+11 = -11\] \[\large x= ??\ \text{ or } \ x = ??\]
mathstudent55
  • mathstudent55
Now you need to do it graphically.
BloomLocke367
  • BloomLocke367
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.
BloomLocke367
  • BloomLocke367
oh my goodness these are getting more and more complex as they go on.

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