x^2 - ( y -6)^2 = 36 and y = -x^2? help please?

PLEASE, no one seems to know how to solve for x.

because this is what i result in y= -72 - y^2

im so confused, i substituted the y for x but i cannot combine the terms. Help please?

5 Answers

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  • 8 years ago

    I think one of the minus sign is wrong, because the equation does not have proper roots.

    Your results are also not correct because

    Substituting y for x^2.

    -y-(y-6)^2=36. Equation cannot be satisfied for any value of y.

  • 8 years ago

    It is so easy dude the technique is in front of you !!!

    Since, y = -(X^2) => (X^2) = -y -[1]

    Substitue in your original equation : (X^2) - (y-6)^2 = 36 -[2]

    Substituting [1] in [2] we get :

    -y - (y-6)^2 = 36

    expand (y-6)^2

    -y - (y^2 - 12y + 36 ) = 36

    -y - y^2 + 12 y - 36 = 36

    -y^2 + 11y = 72

    y^2 - 11y + 72 = 0

    Use quadratic formula to find y and then substitue in (y)^(1/2)

  • 8 years ago

    x^2 - ( y -6)^2 = 36

    x^2 -( -x^2 - 6)^2 = 36

    x^2 -(x^4 + 12x^2 + 36) = 36

    -x^4 - 11x^2 - 36 = 36

    -x^4 - 11x^2 = 72

    x^4 + 11x^2 = 72

    x^4 + 11x^2 + (11/2)^2 = 72 + (11/2)^2

    x^4 + 11x^2 + 5.5^2 = 72 + 5.5^2

    (x^2 + 5.5)^2 = 102.25

    x^2 + 5.5 = ±√102.25

    x^2 = -5.5 ±√102.25

    x = ±√( -5.5 ±√102.25)

  • 8 years ago

    Ok basically what u have to to is expand the bracket so u get

    ,x^2-y^2-12y+36=36

    Now u cut 36 from both side of equation

    U sub in y in terms of x nd work it out

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  • 8 years ago

    x² - (y - 6)² = 36

    x² - (y² - 12y + 36) = 36

    x² - y² + 12y - 36 = 36

    x² - y² + 12y - 72 = 0 → but you know that: y = - x² → x² = - y

    - y - y² + 12y - 72 = 0

    - y² + 11y - 72 = 0 → you multiply by - 1

    y² - 11y + 72 = 0

    Polynomial like: ax² + bx + c, where:

    a = 1

    b = - 11

    c = 72

    Δ = b² - 4ac (discriminant)

    Δ = (- 11)² - 4(1 * 72) = 121 - 288 = - 167 → Δ < 0 → no solution

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