SOLUTION: why dose {{{sqrt(4x^4+36x+81)}}} ={{{2x^2+9}}} and not ={{{2x^2+6x+9}}} what happens to the 6??

Question 245988: why dose
=
and not
=
what happens to the 6??

Found 3 solutions by oberobic, richwmiller, Earlsdon:
Answer by oberobic(2304)   (Show Source): You can put this solution on YOUR website!
When you factor you find that it =

.
To check, multiply through:
=

.
We observe that:
is equivalent to:

.
So, equals:

.
Note: I had to assume the original equation is as I wrote it, which has a in the middle, not .
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
(2x^2+9)^2=(2x^2+9)(2x^2+9)
use FOIL
first
2x^2*2x^2=4x^4 no,problem here
outer
2x^2*9
inner
9*2x^2
last
9*9=81 no problem here
add 2x^2*9and 9*2x^2
18x^2+18x^2=36x^2
so we have 4x^4+36x^2+81 not 36x

{2x^2+6x+9}^2 would be 4x^4+24x^3+72x^2+108x+81
neither work out to be 4x^4+36x+81
I suspect that you copied the original sqrt incorrectly ie sqrt(4x^4+36x+81)

Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
Why does:
and not
Well, you can answer this question by squaring and see what you get. Let's do that just for fun!
Simplify.

Now we'll start with the original expression and factor the radicand:
Take the square root of the right side.

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