Can someone please double check my answer?
Solve the system by method of substitution.
y=x
y=x^3+4x^2+6x
Answer choices:
-(0,0)
-No real solution
I think it is (0,0) but would like to double check

Question

shadowfiend · Answer

You don't even *really* need the substitution method, but if you do use it you end up with:

$$y = y^3 + 4y^2 + 6y$$
$$0 = y^3 + 4y^2 + 5y$$
$$0 = y(y^2 + 4y + 5y)$$

This means that either y must be 0 or $y^2 + 4y + 5y$ must be 0.

shadowfiend · Answer

Sorry, that was $y^2 + 4y + 5$ must be 0, and the equation should be $0 = y(y^2 + 4y + 5)$.

If y = 0 and y = x, then x = 0. So (0, 0) is *a* solution. If you try to solve $y^2 + 4y + 5$ and find where it is 0, you'll find it doesn't have a real solution (the discriminant $b^2 - 4ac$ will be <0), so the only real solution is (0, 0).

shadowfiend · Answer

Hope that helps!

anonymous · Answer

Thanks now what if the equation was:
y=-x
y=x^3+x^2+5x
would it be (0,0) or no real solution? I was thinking no real solution....

shadowfiend · Answer

Well, what do you get if you plug into the bottom equation?

anonymous · Answer

well i thought that because you can't have x=-0... that's just not possible...Right?

shadowfiend · Answer

Oh, sure it is. -0 is the same as 0.

anonymous · Answer

oh ok so it's (0,0) again cool thanks