solve this quadratic system:
4x^2 + 9y^2 = 72
x - y^2 = -1
explain please?

Question

amistre64 · Answer

they want to know what points they have in common

amistre64 · Answer

-72 from each side of the top one

and +1 to both sides of the bottom one 

and you will have 2 equations equal to 0

amistre64 · Answer

since both equation = 0; they equal each other; so make them equal and subtract one equation from the other

amistre64 · Answer

4x^2 + 9y^2 -72 = x - y^2 +1

4x^2 + 9y^2 -72
  -x   + y^2    -1
----------------
4x^2 -x +10y^2 -73 = 0  but i might be trying to do this the ahard way

amistre64 · Answer

try this:

4x^2 + 9y^2 = 72 
      x - y^2 = -1  (*9)

4x^2 + 9y^2 = 72 
      9x - 9y^2 = -9
-----------------
4x^2 +9x = 63

4x^2 =9x -63 = 0 ; solve the quadratic now

amistre64 · Answer

4x^2 +9x -63 = 0 

x = -9/8 + sqrt(81 -4(4)(63))/8

x = -9/8 - sqrt(81 -4(4)(63))/8

amistre64 · Answer

x = -9/8 +- sqrt(1089)/8

amistre64 · Answer

x = -9 +- 33
      ---------
            8

amistre64 · Answer

x = 24/8 = 3

x = -42/8 = -21/4 = -5.25

amistre64 · Answer

plug these into one of the equations to solve for y

anonymous · Answer

For what it's worth, you were (in my opinion) doing it the hard way (not that it makes much difference):

$$x - y^2 = -1 \implies y^2 = x+1$$
Which we can then use in the first equation:
$$4x^2+9(x+1)=72 \implies ... $$

amistre64 · Answer

-72+9 = -63 .... yay!! .. validation  ;)

anonymous · Answer

thnx guys

anonymous · Answer

Note when finding all solutions
$$y^2 = x+1 \implies y=\pm \sqrt{x+1}$$ so it will lead to 4 (x,y) pairs.

amistre64 · Answer

yeah, that first bit was definetly going down the wrong path... lol

amistre64 · Answer

a parabola joined to an ellipse i think would produce at most 2 real points. right?

amistre64 · Answer

nah; 4 is right lol

anonymous · Answer

(Maybe they aren't all real, hehe!)

amistre64 · Answer

i saw it wrong there for a moment :)