Question

How do you solve this?

Answer 1

$Solve\ x^4-13x^2=-36.$

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.

Answer 2

X⁴-13x²+36
(x²-4)(x²-9)



x²=4. | x²=9
x=√4. |x=√9
x=2. |x=3




27x³-8=0
x³=8/27
x=2/3





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