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March 26, 2017

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16x^2+16y^2-16x+24y-3=0
How do I solve this?

  • Algebra 2 - ,

    "Solving" means finding ordered pairs that satisfy the equation.
    Your equation is a circle, and there are an infinite number of ordered pairs that will work.

    Did you mean to put it in the standard form of a circle equation?

  • Algebra 2 - ,

    yes

  • Algebra 2 - ,

    16x^2+16y^2-16x+24y-3=0
    16(x^2 - x + .....) + 16(y^2 + (3/2)y + ... = 3
    16(x^2 - x + 1/4) + 16(y^2 + (3/2)y + 9/16) = 3 + 4 + 9
    16(x-1/2)^2 + 16(y+3/4) = 16
    (x-1/2)^2 + (y+3/4)^2 = 1

    all vital info about the circle is now obvious.

  • Algebra 2 - ,

    thank you very much

  • Algebra 2 - ,

    The coordinate of all relative maxima 2x^4+x^3-33x^2-16x+16

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