Y=2x^2 y^2=x^2+6x+9 what is a possible solution for x in the system of equations above?

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)