Question

i see this doesn't have an x in the 3rd term. I cant factor it that way, then?

solution to this x^4-12x^2+27=0

Answer 1

@jim_thompson5910

Answer 2

\[\large\rm x^4-12x^2+27=0\]Apply exponent rule for this first step,\[\large\rm (x^2)^2-12x^2+27=0\]

Answer 3

ok, got it.

Answer 4

Notice we have some x^2's...
Maybe we can make a substitution to simplify what we're looking at.
Let's replace x^2 with u.
\(\large\rm \color{orangered}{u=x^2}\)

Then our expression,\[\large\rm (\color{orangered}{x^2})^2-12\color{orangered}{x^2}+27=0\]becomes\[\large\rm \color{orangered}{u}^2-12\color{orangered}{u}+27=0\]

Answer 5

And from here, we have a quadratic that can be factored without too much trouble. What do you think about the u business? Confused? :)

Answer 6

nope, completely understand this!!! Now, i have (u-9)(u-3)

Answer 7

Ok great, now undo the substitution that we made,\[\large\rm (u-9)(u-3)=0\]becomes\[\large\rm (x^2-9)(x^2-3)=0\]

Answer 8

ok so re-replace the u with the x^2 that we originally replaced it with and then now i have to solve the parenthesis right.?

Answer 9

Yes :)

Answer 10

You can apply your Zero-Factor property,\[\large\rm (x^2-9)=0\]\[\large\rm (x^2-3)=0\]And solve for x in each case.
Remember, we started with an x to the `fourth power`.
So we expect to end up with `four solution`.

Answer 11

oh COOL. i never knew that!

Answer 12

3,-3
and
sqrt3,-sqrt3

ah yes, good job \c:/

Answer 13

Seminole State? oo that's fun.
I been attending UCF myself c:

Answer 14

say i had (x^2-2), would it be +-sqrt2 in that case?

Answer 15

yeah SSC. Hopefully going straight to UCF afterwards.

Answer 16

Yes, you could either factor it using difference of squares,\[\large\rm x^2-2=0\qquad\to\qquad (x-\sqrt2)(x+\sqrt2)\]Or solving directly, adding 2 to each side,\[\large\rm x^2=2\]and then square rooting,\[\large\rm \sqrt{x^2}=\sqrt2\]Whenever you take the square root of a square x,
that's when the plus/minus shows up,\[\large\rm \pm x=\sqrt2\]And then of course you can move it to the other side,\[\large\rm x=\pm\sqrt2\]