x^2+x^4-20=0
(x^2-4)(x^2+5)=0
(x+2)(x-2)(x^2+5)=0
x= +-2, x= +-sqrt(-5
(2,4)(-2,4)(sqrt-5,-5)(-sqrt-5, -5)
Hi,
I can't seem to figure out the answer to this problem.
Thanks to anyone who can help!
*Solve the system of equations. Write your solution(s) in the form (x,y).
y = x^2
x^2 + y^2 = 20
3 answers
y + y^2 = 20
y^2 + y - 20 = 0
(y-4)(y+5) = 0
y = 4 or y = -5
y = 4 is easy, then x = -2 or +2
y = -5 means x = i sqrt 5
now go back and see which of those solutions work in the original
if y = 4 and x = 2 then
yes, y = x^2
and
yes, 4 + 16 = 20
if y = 4 and x = -2 then
yes, y = x^2
yes, 4 + 16 = 20
if y = -5 and x = i sqrt 5
yes, y = x^2
yes, -5 + 25 = 20
y^2 + y - 20 = 0
(y-4)(y+5) = 0
y = 4 or y = -5
y = 4 is easy, then x = -2 or +2
y = -5 means x = i sqrt 5
now go back and see which of those solutions work in the original
if y = 4 and x = 2 then
yes, y = x^2
and
yes, 4 + 16 = 20
if y = 4 and x = -2 then
yes, y = x^2
yes, 4 + 16 = 20
if y = -5 and x = i sqrt 5
yes, y = x^2
yes, -5 + 25 = 20
I left out y = -5, x = -i sqrt 5
y = (-isqrt5)^2 = -5 yes
-5 + 25 = 20 yes
y = (-isqrt5)^2 = -5 yes
-5 + 25 = 20 yes
0 answers