Question

x^2+y^2=20

y=x^2




A. {(4, -2), (4, 2)
B. {(-2, 4), (2, 4)
C.(2, 4)
D.(3, 9)

Answer 1

X^2+Y^2=20

Answer 2

Y=X^2

Answer 3

i think for this one easiest to check answer and see which ones work

Answer 4

i DONT KNOW HOW TO DO IT

Answer 5

\( x^2+y^2=20 \) and \( y=x^2 \) \( \implies x^4+x^2=20 \)

Answer 6

y^2 + y - 20 = 0 after substitution ,
(y+5)(y-4) = 0

Answer 7

\[ (-2+x) (2+x) \left(5+x^2\right) =0 \]

Answer 8

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)

Answer 9

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

Answer 10

Yes substitution is the best choice here.

Answer 11

now try
\[(-2)^2=4\] true and
\[(-2)^2+4^2=20\] also true

Answer 12

{ (sqrt 5, 5) ( -sqrt 5 , 5 ) ( 2, 4) (-2, 4) }

Answer 13

Analytically it would be messy.

Answer 14

(3,9) does not work because although
\[3^2=9\] is true
\[3^2+9^2\neq 20\] so that one is out

Answer 15

the 5's dont work, scratch that

Answer 16

and (4,-2) does not work because
\[4^2\neq -2\]