Solving systems of equations.?

Question

Can someone show me how to solve this please? the two equations are y= -2x^2 + 2 and y= 2(x^4 - 2x^2 +1) thanks

Answer 1

ok set the two equations equal to each other
-2x^2 + 2 = 2(x^4 -2x^2 +1)
to make things easier, let x^2=z
-2z + 2 = 2(z^2 -2z +1)
divide both sides by 2
-z + 1 = z^2 -2z +1
add z and subtract 1
0 = z^2 -z
factor out a z
0=z(z-1)
so
z=0
or
z-1=0
plug in x^2 for z
x^2=0
or
x^2-1=0

in the first case
x=0

in the second
x^2-1 can be factored so that
(x-1)(x+1)=0
now
x-1 =0
or
x+1=0

so x=1 or x=-1

so, the three answers are x=0,1,-1

now plug these back in for x and solve for y
when x=0
y=-2(0)^2 + 2
y=0 + 2
y=2
when x=1
y= -2(1)^2 + 2
y= -2+2
y=0
when x=-1
y= -2(-1)^2 + 2
y= -2 + 2
y= 0

so, your solutions, in ordered pairs are
(0 , 1) ; (-1, 0) and (1,0)

Answer 2

Set the two equations equal, then factor .

-2x^2+2 = 2(x^4-2x^2+1)
-2(x^2 -1) = 2(x^2-1)^2
Bring to one side, set equal zero.
2(x^2-1)^2+2(x^2-1) = 0
Factor further:
2(x^2-1) * (x^2-1+1) = 0
Zero property rule:
x=plus or minus one and x=0

Answer 3

y= -2x^2 + 2
y= 2(x^4 - 2x^2 +1)
Simutaneous Equations
y= -2x^2 + 2
y= 2x^4 - 4x^2 +2
-----------------------
0 =-2x^4+2x^2
0 = 2x^4-2x^2
=x^2-1
=(x+1)(x-1)
x = -1 or x = 1

Answer 4

y= -2x^2 + 2 = 2x^4 -4x^2 +2
2x^2=2x^4
1=x^2
x=+1 or -1

Answer 5

y= -2x^2 + 2 and y= 2(x^4 - 2x^2 +1) eliminate y
-2x^2+2=2(x^4-2x^2+1) divide both sides by 2
-x^2+1=x^4-2x^2+1 add x^2-1 to each side
x^4-x^2=0
x^2(x^2-1)=0
x^2(x+1)(x-1)=0
x=0
y=-2x^2+2
y=2
(0, 2)

x=1
y=-2x^2+2
y=-2+2
y=0
(1, 0)

x=-1
y=-2x^2+2
y=-2+2
y=0
(-1, 0)

solutions are: (0, 2); (1, 0) & (-1, 0)

Answer 6

Well, since they are both equal to y, they are both equal to each other so you get
-2x^2+2=2(x^4-2x^2+1)
1-x^2=x^4-2x^2+1
x^4-x^2=0
x^2(x^2-1)=0
x=0,0,1,-1

Answer 7

substitute 2x^2=t

then solve the system for substitute y and solve for t
then for y from 1st equation
then find x=+-sqrt(t/2)