anonymous
  • anonymous
x^2+y^2=20 y=x^2 A. {(4, -2), (4, 2) B. {(-2, 4), (2, 4) C.(2, 4) D.(3, 9)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
X^2+Y^2=20
anonymous
  • anonymous
Y=X^2
anonymous
  • anonymous
i think for this one easiest to check answer and see which ones work

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anonymous
  • anonymous
i DONT KNOW HOW TO DO IT
anonymous
  • anonymous
\( x^2+y^2=20 \) and \( y=x^2 \) \( \implies x^4+x^2=20 \)
perl
  • perl
y^2 + y - 20 = 0 after substitution , (y+5)(y-4) = 0
anonymous
  • anonymous
\[ (-2+x) (2+x) \left(5+x^2\right) =0 \]
lgbasallote
  • lgbasallote
y + y^2 = 20 y^2 + y + 1/4= 20 +1/4 (y+1/2)^2 = 81/4 y + 1/2 = 9/2 y = 4 knowing that... 4 = x^2 x = +/- 2 so it's {(-2,4)(2,4)
anonymous
  • anonymous
for example, try (2,4) \[2^2=4\] so second equation is satisfied, but \[2^2+4^2=4+16=20\] so first equation is satisfied as well
anonymous
  • anonymous
Yes substitution is the best choice here.
anonymous
  • anonymous
now try \[(-2)^2=4\] true and \[(-2)^2+4^2=20\] also true
perl
  • perl
{ (sqrt 5, 5) ( -sqrt 5 , 5 ) ( 2, 4) (-2, 4) }
anonymous
  • anonymous
Analytically it would be messy.
anonymous
  • anonymous
(3,9) does not work because although \[3^2=9\] is true \[3^2+9^2\neq 20\] so that one is out
perl
  • perl
the 5's dont work, scratch that
anonymous
  • anonymous
and (4,-2) does not work because \[4^2\neq -2\]

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