How do you simplify #sqrt(1+9x^4)#?

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Answer 1 · 2015-08-08T18:42:03

That's about as simple is it can be.

You can factor the radicand as

#(3x^2-sqrt(6)x+1)(3x^2+sqrt(6)x+1)#

but there are no square factors to allow simplification.

If the radicand of a square root has a square factor, then you can move that outside the square root.

For example, #sqrt(12) = sqrt(2^2*3) = 2sqrt(3)#

With variables you have to be a little more careful, but you can say:

#sqrt(9x^2) = 3abs(x)#

In the case of #(1+9x^4)# there is no square factor - the #9# and #x^4# do not really help as they are not factors of the whole expression #(1+9x^4)#

We can factor the radicand to get:

#sqrt(1+9x^4) = sqrt((3x^2-sqrt(6)x+1)(3x^2+sqrt(6)x+1))#

but I would not call that simpler.