why doesn't 1 work...it's in the domain of the function and your work looks right...did you write the problem down wrong?

you should put x = 1 for every single line and see at what point it fails to work

Uhh, since I already solved for x, and I got 2/3 or 1, shouldn't the number when plugged in, get the equation equal?
If I plug in 2/3, then I would get the same answer on both sides.
For 1 though, I only have 1 + 11 = 7(1) + 3 and 12 = 10




I reviewed the answers, and correct choice was 6x^2 - 30.
I typed it out twice in 6 and 6x, so I guess the typo led to the mistake there.
Guess you can quote brmv and tell him to refresh it.

Edit: there is actually another choice that is 6x^3 - 30x
If I did it step by step again
g(f(x))
g(x^2 - 5) <--- g(x) technically since F is already x
6(x^2 - 5) = 6x^2 - 30x

So when g(x) = 6x, the x is the function of F rather than just an x.
Unless I'm doing the same mistake but otherwise this should be the final answer.

This post was edited by Sefira on Nov 30 2013 07:10pm

all positive numbers have 2 square roots, so I believe your simplified radical (sqrt(1)) can be either +1 or -1.
If you let the radical be -1, then your equation works.

So you have two possible answers:
1 + 11(1) = 7(1) + 3
or
-1 + 11(1) = 7(1) + 3

How? I might be overlooking something simple to not notice but;
I factored out the 15x^2 - 25x + 10 first, and x - 1 = 0
I didn't get 1 from the radical, and since it's a principal root it can't be -1

Your calculated zeros are correct, but the problem lies in when you plug those values back into the original equation.
(Radical x^2 + x - 1) + 11x = 7x + 3
now plug in 1:
(Radical 1^2 + 1 - 1) + 11(1) = 7(1) + 3
then simplifying, you have:
(Radical 1) + 11 = 10 --> (Radical 1) = -1

here's a quick video that might help explain it clearer than I am

Ah I see, I lost my head or something there, thanks for the help.

no problemo

ah so x=1 is a solution...but when you plug it in, you consider the "-1" answer ...

Final problems (and the hardest)

Express in simplest form:
[ 4 - x^2 / x^2 + 7x + 12] / [2x - 4 / x + 3]
So, I tried factoring out the denominators of both, and simply cancel out the x + 3. After, I would move the 2x - 4 that remains onto the numerator. Is this the correct way to do it?

Solve for x and express in simplest a + bi form: 3x^2 - 6x + 4 = 0
I completely forgot how to do a + bi form. The only thing I remember is using the quadratic formula, if that is correct.

Any tips/advice appreciated ! Eating lunch now, so I'll probably try to do these later as well.