help with algebra?

Question

so i'v tried SO MANY TIMES to get this and i just cant

Solve the inequality
1. |3 - 3x/4| >= 9

2. |3x - 3/5| < 0.2

3. Find the absolute inequality corresponding to x <= -22/3 or x >= 11/3

Solve for x

4. (x - 2)(x + 1) = 4

5. x^3 + 2x^2 - 9x = 18

Factor completely

6. 64 + a^3

7. 2x^4 + 16x

8. 32a^4 - 162b^4

Simplify (be sure to rationalize all denominators)

9. CUBERT(-27)

10. 3b[SQRT(27a^5b)] + 2a[SQRT(3a^3b^3)]

11. 3/[SQRT(5) - 2]

12. x^(3/4) + 5/[x^(1/4)] (you may assume x>0)

Solve the following (be sure to check for extraneous roots)

13. SQRT(4a + 1) = 1

14. CUBERT(3x - 8) = 1

15. SQRT(x + 4) = SQRT(x) - 2

16. (3x + 1)^(1/4) = 2

17. SQRT(y - 8) = SQRT(8 - y)

18. SQRT(4a+7) = -SQRT(a+2)

19. SQRT(x^2 - 5x + 2) = 2x

20. http://i15.tinypic.com/8g74k5e.jpg

PLEASE SHOW WORK!

Answer 1

For starters, I don't do inequalities. It's against my religion. As for the rest, I'll get you started, but I'm not going to do all the work for you. You should have learned enough to do most of these questions with a little help.

For #4, you multiply out the left side, then combine with the 4 to make a quadratic equation =0. Then you factor it. For #5, move the 18 to the left side, find three factors that multiply together to give -18, and try out the possibilities. e.g., (x+2)(x+3)(x-3). If that doesn't work, switch the signs and try(x-2)(x+3)(x+3).

#6 is already completely factored. In #7, you can factor out 2x to give 2x(x^2 + 8). For #8, you should look for the difference of two squares. If you first factor out a 2, you get 16a^4 - 81b^4 left. That should make it obvious. For #9, 3 is the cube root of +27, so -3 should be the cube root of -27. That works with all odd powers.

For 10, factor out any square terms you can find under the two sqrt terms, then move them out of the sqrt area. In both, I found that by factoring out 3ab, you are left with squares that can be removed. In the first step, it would read 3b*sqrt(3ab*9a^4) + 2a*sqrt(3ab*a^2b^2). The next step would read 3b*3a^2*sqrt(3ab) + 2a*ab*sqrt(3ab). Multiply the first part of each term to get 9a^2b*sqrt(3ab) + 2a^2b*sqrt(3ab). You can now add them, because the first factors of each term share common variables. So you would finally get 11a^2b*sqrt(3ab).

In #11, multiply top and bottom by (sqrt5+2). This takes advantage of the "difference of two squares" to get rid of the sqrt5 on the bottom. Sorry, #12 is beyond my expertise. For #13, just square both sides, and for #14, just cube both sides. That removes the root signs, and they become simple equations.

For #15, square both sides and solve the simple equation. For #16, take each side to the fourth power to remove the 1/4 exponent. For #17 and #18, square both sides - you can ignore the minus sign in #18, because -*-=+. For #19, square both sides, move the 4x^2 to the left, multiply through by -1, and solve the quadratic equation by factoring.

I'll leave #20 for someone else. If you can't finish most of the questions with the help I've given you, then I think you should get regular tutoring help, take the course again next semester, or switch to another subject.

Answer 2

Go back a lesson and reread it and the instructions for this lesson. Does your book have practice problems with the answers in the back?

Answer 3

i can help u with solve for x
[x-2][x+1]=4
2x=4+2-1
2x=6-1
2x=5
x=5/2

Answer 4

Dude if I do this for you you won't learn it so go to a friends and ask help from them

Answer 5

Ha! Cheating on your homework!
If I tell you, you will never learn!
Bwahahahahahahahahey wait a minute? when did I get this evil?