|
Title: How does one do systems of quadratics? Post by: !JRon@n on Aug 28, 2012 I'm pretty lost. Here's an example problem. Please show and explain steps. Thank you.
4x^2+9y^2=72 x-y^2=-1 Title: How does one do systems of quadratics? Post by: rjframe on Aug 28, 2012 Content hidden
Title: How does one do systems of quadratics? Post by: MikeLake on Aug 28, 2012 Problems of this type which involves different degree of variables is done using elimination method
4x^2+9y^2=72 --------------------------1) x-y^2=-1-------------------------------------2) Noticing that both first and second equation contain second degree y variable we will eliminate that first from 2 we have y^2 = x +1 substituting that in 1 we have 4x^2 + 9(x+1) = 72 4x^2 +9x+9 = 72 4x^2+9x -63 = 0 so x = -b +/= sqrt(b^2 - 4ac) /2a x = - 9 +/- sqrt( 81 - 4*4*(-9) / 8 = -9+/- 33 / 8 = 3 or -21/4 y^2 = x +1 so y^2 = 3 +1 or -21/4 +1 y^2 = 4 or -17/4 | y becomes imaginary neglecting that value we have (3,2) or (3.-2) as answer you can substitute and check the answer |