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