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bmorg980

  • 2 years ago

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

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  1. sasogeek
    • 2 years ago
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    have you tried solving this?

  2. bmorg980
    • 2 years ago
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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?

  3. sasogeek
    • 2 years ago
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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?

  4. bmorg980
    • 2 years ago
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    yes! please

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

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

  7. sasogeek
    • 2 years ago
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    sorry for late reply, i lost connection

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

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

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

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

  12. bmorg980
    • 2 years ago
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    oops, i subtracted instead of adding. so 1?

  13. sasogeek
    • 2 years ago
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    yeah :)

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

  15. sasogeek
    • 2 years ago
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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 \(\huge (x+\frac{1}{2})^2=1\) :)

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

  17. sasogeek
    • 2 years ago
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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 :)

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