## anonymous 5 years ago solve this quadratic system: 4x^2 + 9y^2 = 72 x - y^2 = -1 explain please?

1. amistre64

they want to know what points they have in common

2. amistre64

-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

3. amistre64

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

4. amistre64

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

5. amistre64

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

6. amistre64

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

7. amistre64

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

8. amistre64

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

9. amistre64

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

10. amistre64

plug these into one of the equations to solve for y

11. anonymous

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: $4x^2+9(x+1)=72 \implies ...$

12. amistre64

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

13. anonymous

thnx guys

14. anonymous

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

15. amistre64

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

16. amistre64

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

17. amistre64

nah; 4 is right lol

18. anonymous

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

19. amistre64

i saw it wrong there for a moment :)