Respuesta :
Answer:
So we have the two real points (3/2 , 9/2) and (-1,2).
(Question: are you wanting to use the possible rational zero theorem? Please let me know if I didn't answer your question.)
Step-by-step explanation:
y=2x^2
y^2=x^2+6x+9
is the given system.
So my plain here is to look at y=2x^2 and just plug it into the other equation where y is.
(2x^2)^2=x^2+6x+9
(2x^2)(2x^2)=x^2+6x+9
4x^4=x^2+6x+9
I'm going to put everything on one side.
Subtract (x^2+6x+9) on both sides.
4x^4-x^2-6x-9=0
Let's see if some possible rational zeros will work.
Let' try x=-1.
4-1+6-9=3+(-3)=0.
x=-1 works.
To find the other factor of 4x^4-x^2-6x-9 given x+1 is a factor, I'm going to use synthetic division.
-1 | 4 0 -1 -6 -9
| -4 4 -3 9
|________________ I put that 0 in there because we are missing x^3
4 -4 3 -9 0
The the other factor is 4x^3-4x^2+3x-9.
1 is obviously not going to make that 0.
Plug in -3 it gives you 4(-3)^3-4(-3)^2+3(-3)-9=-162 (not 0)
Plug in 3 gives you 4(-3)^3-4(-3)^2+3(-3)-9=72 (not 0)
Plug in 3/2 gives you 4(3/2)^2-4(3/2)^2+3(3/2)-9=0 so x=3/2 works as a solution.
Now let's find another factor
3/2 | 4 -4 3 -9
| 6 3 9
|________________________
4 2 6 0
So we have 4x^2+2x+6=0.
The discriminant is b^2-4ac which in this case is (2)^2-4(4)(6). Simplifying this gives us (2)^2-4(4)(6)=4-16(6)=4-96=-92. This is negative number which means the other 2 solutions are complex (not real).
So the other real solutions that satisfy the system is for x=3/2 or x=-1.
Since y=2x^2 then for x=3/2 we have y=2(3/2)^2=2(9/4)=9/2 and for x=-1 we have y=2(1)^2=2.
So we have the two real points (3/2 , 9/2) and (-1,2)
