I already did the work, hoping some people could look over it. Just some basic Algebra II, but I'd rather not make any simple mistakes.

1. (1 - 4/x) / 1 - 2/x - 8/x^2
Multiply both by x^2 to remove denominators.
x^2 - 4x / x^2 - 2x - 8
x(x-4) / (x-4)(x+2) = x / x+2

2. f(x) = x^2 - 5 and g(x) = 6x, then g(f(x)) is equal to
g(x^2-5) = 6(x^2-5) = 6x^2 - 30x

3. What is the range of f(x) = (x+4)^2 + 7
I put y => 7 since whatever the domain is, it's still being squared and it will never be a negative, and 7 remains a constant.

4. Solve the equation 8x^3 + 4x^2 - 18x - 9 = 0 algebraically for all values of x.
4x^2(2x+1) - 9(2x+1) = 0
(4x^2-9)(2x+1) = 0
x^2 = 9/4 x = 3/2
x = -1/2

5. The roots of the equation 9x^2 + 3x - 4 = 0 are
b^2 - 4ac
(3)^2 - 4(9)(-4) = 9 + 144 = 153.
The roots are real, irrational, and unequal

1. (1 - 4/x) / 1 - 2/x - 8/x^2
Multiply both by x^2 to remove denominators.
x^2 - 4x / x^2 - 2x - 8
x(x-4) / (x-4)(x+2) = x / x+2
good but if you might want to add x≠0, x≠4

2. f(x) = x^2 - 5 and g(x) = 6x, then g(f(x)) is equal to
g(x^2-5) = 6(x^2-5) = 6x^2 - 30x
good


3. What is the range of f(x) = (x+4)^2 + 7
I put y => 7 since whatever the domain is, it's still being squared and it will never be a negative, and 7 remains a constant
good.

4. Solve the equation 8x^3 + 4x^2 - 18x - 9 = 0 algebraically for all values of x.
4x^2(2x+1) - 9(2x+1) = 0
(4x^2-9)(2x+1) = 0
x^2 = 9/4 x = +/- 3/2
I added a +/-
x = -1/2

5. The roots of the equation 9x^2 + 3x - 4 = 0 are
b^2 - 4ac
(3)^2 - 4(9)(-4) = 9 + 144 = 153.
The roots are real, irrational, and unequal
good

This post was edited by Rocinante on Nov 27 2013 06:49pm

there must be a typo in 1, "(1 - 4/x) / 1 - 2/x - 8/x^2 " does not make sense without '=' and some parantheses/brackets

2 and 3 are correct
might look at 4 and 5 later

you should double-check yourself using a different method for each one. you learn much better that way.


Use some value of x in the domain of both expressions (eg: 1) and make sure they yield the same result. you can test this yourself.

again, you can test this yourself. pick some value of x in the domain of f and calculate f(x). then use that value in g(f(x)). compare that with your equation.

graph the function to verify.


put all values you find back into the equation. should be 0. if you get fewer solutions than the degree of the polynomial and it's not obvious why, then graph it to verify.


graph and see if it ever intercepts the axis.

This post was edited by carteblanche on Nov 27 2013 06:52pm

Oh, that was just asking for the simplest form.


Thanks, we haven't done graphing yet, I'll try to look in the book to see how I should do it.



Thanks a lot.

This post was edited by Sefira on Nov 27 2013 06:54pm

#2 is 6x^(2) - 30. I'm not sure how the 30x got in there.

#4 can be furthur simplified to (2x + 3)(2x - 3)(2x +1) = 0
giving you roots of x=(3/2), x=(-3/2), and x=(-1/2)

The rest look correct, good job

not much graphing to be done
since it must be a vertical parabola (the x^2 being the highest power and has a positive coefficent)
now
x=0 -> y=-4
x=1 -> y=8
x=-1 -> y=2
so the axis is crossed in each of the intervals [-1,0] and [0,1]



you better refresh how g(f(x)) works

This post was edited by brmv on Nov 27 2013 07:12pm

Thanks for the tip, Rocinante mentioned +/- but I guess I'll also factor it out further to show the work.

For #2 I believe you just put 6x(x^2-5), since x^2-5 is part of the whole x in terms of G.



I see, I've only really remembered how the parabolas are from basic Algebra, but I think I'll be learning graphing them along with logarithms after Thanksgiving.

This post was edited by Sefira on Nov 27 2013 07:17pm

Have another problem.

(Radical x^2 + x - 1) + 11x = 7x + 3
(Radical x^2 + x - 1) = -4x + 3
(Radical x^2 + x -1)^2 = (-4x + 3)^2
x^2 + x -1 = (-4x + 3)(-4x + 3)
x^2 +x - 1 = 16x^2 -24x + 9
0 = 15x^2 - 25x + 10

This part got me. I couldn't factor it out using just binomials. So instead I did
0 = 5(3x^2 - 5 + 2)
0 = 5(3x - 2)(x -1)
3x - 2, x - 1 = 0
x = 2/3 / 1
For x = 2/3, I get 7.667 = 7.667, so this works.
For x = 1, I got 12 =/= 10, so that doesn't work.

Just wondering if there's any mistakes in my work and if my answer is correct or I just got lucky..

This post was edited by Sefira on Nov 30 2013 06:34pm

I still do not see how it comes out to be 6x^(2) - 30x.
I'm getting 6x^(2) - 30

I must be missing something blatantly obvious, can someone work it out step-by-step?