[tex]Solve\ x^4-13x^2=-36.[/tex]
View this equation as a quadratic like so:
[tex]x^2(x^2)-13x(x)+36=0[/tex]
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:
[tex]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[/tex]
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.
[tex]x^2-4=0,\ or\ x^2-9=0[/tex]
We can add that constant to the right side...
[tex]x^2=4,\ or\ x^2=9[/tex]
And take the square root of each side.
[tex]\boxed{x=\pm2\ or\ \pm3}[/tex]
So, your possible x values are
2, -2, 3, and -3.
