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

2 Answers

Answer:

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.

Sep 24, 2015

Answer:

#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#.