Question

How do you solve `x^4-18x^2+81=0`?

Answer 1

Refer to explanation

Explanation:

It is easy to see that

`x^4-18x^2+81=(x^2)^2-2*9*x^2+9^2=0=>(x^2-9)^2=0`

Hence we have that `(x^2-9)^2=0=>x^2-9=0=>x=3 or x=-3`

Be aware that roots `x_1=3,x_2=-3` have multiplicity of `2`
because we have a fourth degree polynomial.

Answer 2

`x = +-3`

Explanation:

Normally, to solve a polynomial of degree 4 like the one here, you need to do synthetic division and use a lot of theorems and rules - it gets kinda messy. However, this one is special because we can actually make it a quadratic equation.

We do this by letting `u = x^2`. Don't worry about where `u` came from; it's just something we're using to simplify the problem. With `u = x^2`, the problem becomes
`u^2-18u+81 = 0`.

Doesn't that look better? Now we're dealing with a nice, easy quadratic equation. In fact, this is a perfect square; in other words, when you factor it, you get `(u-9)^2`. Of course, we could use the quadratic formula or completing the square to solve this equation, but you're usually not lucky enough to have a perfect square quadratic - so take advantage. At this point, we have:
`(u-9)^2 = 0`

To solve, we take the square root of both sides:
`sqrt((u-9)^2) = sqrt(0)`
And this simplifies to
`u-9 = 0`

Finally, we add 9 to both sides to get
`u = 9`

Awesome! Almost there. However, our original problem has `x`s in it and our answer has a `u` in it. We need to convert `u = 9` into `x =` something. But have no fear! Remember at the beginning we said let `u = x^2`? Well now that we have our `u`, we just plug it back in to find our `x`. So,
`u = x^2`
`9 = x^2`
`sqrt(9) = x`
`x = +-3` (because `(-3)^2 = 9` and `(3)^2 = 9`)
Therefore, our solutions are `x = 3` and `x = -3`. Note that `x = 3` and `x = -3` are double roots, so technically, all of the roots are `x = 3`, `x = 3`, `x = -3`, `x = -3`.