Finding the point of intersection of $y=x^4$ and $x=y^2$.
I need to find the point where $y=x^4$ and $x=y^2$ intersect. Since $x=y^2$ is not in $y$, I need to square root both sides and get $\sqrt{x}$. Where I then set both of them equal to each other. $x^4=\sqrt{x}$. I then move $\sqrt{x}$ to the other side and get $x^4-\sqrt{x}=0$, but this is where I'm stuck. What I did is correct, right? and how would I find the point of intersection. I know one of them is $0$ since that is the origin of both functions.
$\endgroup$ 14 Answers
$\begingroup$Observe that the equations imply that $y=(y^2)^4=y^8$ or equivalently $y(y^7-1)=0$ so that $y=0$ or $y=1$. You can substitute to find $x$.
$\endgroup$ $\begingroup$In addition to other answers, you may find complex answers:
$$y=x^2$$
Substitute $x=y^4$ in.
$$y=(y^4)^2=y^8$$
$$y^8-y=0$$
$$y(y^7-1)=0$$
We have two solutions here:
$$y=0$$
Or
$$y^7-1=0$$
$$y^7=1$$
For complex solutions:
$$y=e^{\frac{2n}7\pi i},n=0,1,2,3,4,5,6$$
Plugging this back into the first equation, we have:
$$x=0,y=0$$
$$x=e^{\frac{n}7\pi i},y=e^{\frac{2n}7\pi i},n=0,1,2,3,4,5,6$$
$\endgroup$ $\begingroup$If you know that $x=y^2$ then just substitute x in the other equation. You get $y=y^8$:the solutions are $y=0$ and $y=1$. If $y=0$ you get $x=0$, if $y=1$ you get $x^2=1$ then $x=1$ or $x=-1$. The solutions are then the $(x,y)$ couples $(0,0),(1,1),(-1,1)$.
$\endgroup$ $\begingroup$From where you left off:
$x^4-\sqrt x=0$
Now factor out a $\sqrt x$
$\sqrt x (x^{\frac 72}-1)=0$
Now you can set
$\sqrt x = 0 \implies x =0$
$x^{\frac 72}-1=0 \implies x=1$
Now that you have your $x$ coordinates, you can plus them in in either equation and find the $y$ coordinates.
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