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bmorg980

i need to complete the square of -36x^2-36x+27

  • one year ago
  • one year ago

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  1. sasogeek
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    have you tried solving this?

    • one year ago
  2. bmorg980
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    yes. initially i divided each variable by 36, but then i got a bunch of fractions and it did not make sense to me. Do i need to divide by 36?

    • one year ago
  3. sasogeek
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    ok well that's a good start :) and yes you will end up with a bunch of fractions, it happens ;) i'll tell you what to do, try that, then let me know what answer you get... are we good?

    • one year ago
  4. bmorg980
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    yes! please

    • one year ago
  5. bmorg980
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    ok, i got x^2+x+(1/2)^2=-27/36-1/4

    • one year ago
  6. sasogeek
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    uhhh close, but not quite. you should've divided by -36 that way you'd have \(\large x^2+x=\frac{27}{36}\)

    • one year ago
  7. sasogeek
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    sorry for late reply, i lost connection

    • one year ago
  8. bmorg980
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    ok, im going to try that. no problem, im doing some other problems while i wait

    • one year ago
  9. sasogeek
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    then with the next step, you'd get \(\large x^2+x+\frac{1}{4}=\frac{27}{36}+\frac{1}{4} \)

    • one year ago
  10. bmorg980
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    ok, i have x^2+x+1/4=1/2

    • one year ago
  11. sasogeek
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    \(\large \frac{27}{36}+\frac{1}{4}= what? \)

    • one year ago
  12. bmorg980
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    oops, i subtracted instead of adding. so 1?

    • one year ago
  13. sasogeek
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    yeah :)

    • one year ago
  14. bmorg980
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    ok, now im a little confused, isnt it (x+b/2)^2. i dont know where to go from here

    • one year ago
  15. sasogeek
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    ok so you have \(\huge x^2+x+\frac{1}{4}=1 \) what you said is right, b=1, hence the next step is \(\huge (x+\frac{1}{2})^2=1\) :)

    • one year ago
  16. bmorg980
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    so the answer is 1-(x+(1/2))^2

    • one year ago
  17. sasogeek
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    no... i keep losing connection so bear with the reply timing :) what you do is solve for x.... the next step is to find the square root of both sides of the equation to clear the square on the left side... \(\huge x+\frac{1}{2}= \pm \sqrt{1}\) \(\huge x= \pm \sqrt{1}-\frac{1}{2} \) \(\huge x= \pm 1-\frac{1}{2} \) solve for x now :)

    • one year ago
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