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steve816

  • one year ago

Circle question. How would I convert this to standard form?

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  1. steve816
    • one year ago
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    \[x^2+y^2+4x-4y-1=0\]

  2. steve816
    • one year ago
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    @jim_thompson5910 Do you think you can help me with this?

  3. jim_thompson5910
    • one year ago
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    focus on just x^2 + 4x what's missing to complete the square?

  4. steve816
    • one year ago
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    Oh, so we have to complete the square for both x and y?

  5. jim_thompson5910
    • one year ago
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    correct

  6. steve816
    • one year ago
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    for x it would be 4 and for y, it would also be 4 I think.

  7. jim_thompson5910
    • one year ago
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    yeah x^2 + 4x + 4 is a perfect square

  8. jim_thompson5910
    • one year ago
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    so you have to add 4 to both sides to go from x^2 + 4x to x^2 + 4x + 4

  9. steve816
    • one year ago
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    Yes, and that can be simplified to (x+2)^2

  10. steve816
    • one year ago
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    I think I'm understanding this!

  11. jim_thompson5910
    • one year ago
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    yep x^2 + 4x + 4 factors to (x+2)^2

  12. steve816
    • one year ago
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    I did the math and I got (x+2)^2+(y-2)^2=1

  13. jim_thompson5910
    • one year ago
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    you forgot to add 4 to both sides. You only added to the left side

  14. jim_thompson5910
    • one year ago
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    you're adding 2 copies of 4 actually

  15. steve816
    • one year ago
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    oh OOPS I forgot!

  16. steve816
    • one year ago
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    so it would be (x+2)^2+(y-2)^2=9

  17. jim_thompson5910
    • one year ago
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    bingo

  18. steve816
    • one year ago
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    Thank you :D

  19. jim_thompson5910
    • one year ago
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    no problem

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