Question 115431: we have been studying systems of nonlinear equations in two variables.
Question:
Solve: x^2-y^2=1 and x^2+y=7
I started by isolating the y variable on the 2nd problem
y=7-x^2
Then I know I need to use addition or substitution method to solve:
I picked substitution method.
x^2-(7-x^2)^2=1
x^2 -(49-14x^2+x^4)=1
x^2-49+14x^2-x^4=1 changed signs because of negative sign.
-x^4+15x^2-49=1 combined like terms
-1 -1
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-x^4+15x^2-50=0
0=x^4-15x^2+50 rewrite for better understanding
A=x^2 Use this in replacement of x^2
0=A^2-15A+50
(A-10) (A-5) Factor
A=10 A=5
Now I place x^2 back in for A
x^2=10 x^2=5
This is where I get lost, I know on old problems I would use my radicand/square root symbol
over each and x would equal a square root ie: x^2=1 x^2=9
x= +/-1 x= +/-3
and my answers were: (1,3) (3,1)
(-1,-3) (-3,-1)
Please help me finish this problem:) It's been 2 days trying to figure out what I'm doing wrong.
Answer by vertciel(183)   (Show Source):
You can put this solution on YOUR website! Hello there,
1. x^2 - y^2 = 1 and x^2 + y = 7
The easier method is actually elimination.
If I subtract both equations, I get:
-y^2 - y = -6
-y^2 - y + 6 = 0
-(y^2 + y - 6)= 0
-(y + 3)(y - 2) = 0
y = -3, 2
Can you take it from here?
2. -x^4 + 15x^2 - 50 = 0
Let y = x^2.
-y^2 + 15y - 50 = 0
y = 5, 10
Since y = x^2, you have the following two equations:
x^2 = 5 and x^2 = 10
If you square root both sides, you get  and  . If you want to check them, plug them into the original equation.
Hope this helps! Write back for more help!
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