![Solve\ x^4-13x^2=-36.](https://tex.z-dn.net/?f=Solve%5C%20x%5E4-13x%5E2%3D-36.)
View this equation as a quadratic like so:
![x^2(x^2)-13x(x)+36=0](https://tex.z-dn.net/?f=x%5E2%28x%5E2%29-13x%28x%29%2B36%3D0)
We can factor just like a normal quadartic!
We want two numbers that multiply to 36x² and add to -13x.
These numbers are -9x and -4x.
Because our leading coefficient is 1, we can factor straight to (x²-4)(x²-9).
Here's what it would look like if we split the middle and factored:
![x^4-9x^2-4x^2+36=0\\x^2(x^2-9)-4(x^2-9)=0\\(x^2-4)(x^2-9)=0](https://tex.z-dn.net/?f=x%5E4-9x%5E2-4x%5E2%2B36%3D0%5C%5Cx%5E2%28x%5E2-9%29-4%28x%5E2-9%29%3D0%5C%5C%28x%5E2-4%29%28x%5E2-9%29%3D0)
Of course, any value which causes either factor to equal zero is a solution.
The other factor times zero is still going to be zero, of course.
Let's derive these two possibilities from our equation.
![x^2-4=0,\ or\ x^2-9=0](https://tex.z-dn.net/?f=x%5E2-4%3D0%2C%5C%20or%5C%20x%5E2-9%3D0)
We can add that constant to the right side...
![x^2=4,\ or\ x^2=9](https://tex.z-dn.net/?f=x%5E2%3D4%2C%5C%20or%5C%20x%5E2%3D9)
And take the square root of each side.
![\boxed{x=\pm2\ or\ \pm3}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D%5Cpm2%5C%20or%5C%20%5Cpm3%7D)
So, your possible x values are
2, -2, 3, and -3.