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

Refer to explanation

#### Explanation:

It is easy to see that

${x}^{4} - 18 {x}^{2} + 81 = {\left({x}^{2}\right)}^{2} - 2 \cdot 9 \cdot {x}^{2} + {9}^{2} = 0 \implies {\left({x}^{2} - 9\right)}^{2} = 0$

Hence we have that ${\left({x}^{2} - 9\right)}^{2} = 0 \implies {x}^{2} - 9 = 0 \implies x = 3 \mathmr{and} x = - 3$

Be aware that roots ${x}_{1} = 3 , {x}_{2} = - 3$ have multiplicity of $2$
because we have a fourth degree polynomial.

Sep 24, 2015

$x = \pm 3$

#### 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
${u}^{2} - 18 u + 81 = 0$.

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 ${\left(u - 9\right)}^{2}$. 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:
${\left(u - 9\right)}^{2} = 0$

To solve, we take the square root of both sides:
$\sqrt{{\left(u - 9\right)}^{2}} = \sqrt{0}$
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,
$u = {x}^{2}$
$9 = {x}^{2}$
$\sqrt{9} = x$
$x = \pm 3$ (because ${\left(- 3\right)}^{2} = 9$ and ${\left(3\right)}^{2} = 9$)
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 $x = 3$, $x = 3$, $x = - 3$, $x = - 3$.