# 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?

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• 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.

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

Since, y = -(X^2) => (X^2) = -y -

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

Substituting  in  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)

• 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)

• 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

• 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