math help?

Question

what's the answer to this: p.s. x2= x to the second power, etc. not x times 2 because that is 2x....

(x2 - 9)(x3 - 8)(2x - 8) = 0 please help. not that great a math. been working on it, can't get it. thanks guys

i mean like what number does x represent?

THANKS GUYS NOT GOING TO PICK A BEST ANSWER THIS TIME, BUT YEAH. SO YEAH. ALL BEST ANSWERS! just pretend the 10 points are there!!!

Answer 1

3, -3, 2, 4

to get it to equal zero, one of the 3 parts of the equation must equal zero (since 0 times anything is 0). so to get the first part to equal zero 3 or -3 would work. for the second part, 2x2x2 is 8, so if x=2 the second part of the equation would be zero. For the third part x must equal 4 to make that section equal zero.

Answer 2

Word Arithmetics - Part 8
© Copyright 2000, Jim Loy

This one is really tough:

A
----
BCD)AEFG
HICJ
----
HAE
Try to solve it before you look at my solution.


--------------------------------------------------------------------------------

My solution:

We can easily list the possible permutations of A, H, D, and J:

x x
A = 2222233333444466677777888899999
H = 1111122222333355566666777788888
D = 3478946789278937923489234923467
J = 6846828147882682441863642287643
Two of those can be eliminated, as H and J cannot be the same. Now an important clue is that AxBCD=HICJ, where H is one less than A. Only certain values of B will work for the above permutations:

x y x
A = 22222x33334444666777x7888899999
H = 11111x22223333555666x6777788888
D = 34789x67892789379234x9234923467
J = 68468x81478826824418x3642287643
B = 55555 76667977988889 8999
76676 98789 98 9 99
87897 99
999
Those are all of the possible values of B which do not conflict with the other four letters. That eliminates the six columns on the right. Also, we see that E is less than H, but is not zero. That eliminates the five columns on the left. It also eliminates the column that I marked "y" above. We can list the above permutations, adding the letters E and G to the list:

A = 33 333 33 44 4 44 44 6 66 6 77 77 7 7 8 8 8
H = 22 222 22 33 3 33 33 5 55 5 66 66 6 6 7 7 7
D = 66 888 99 22 7 88 99 3 77 9 22 33 4 9 2 3 4
J = 88 444 77 88 8 22 66 8 22 4 44 11 8 3 6 4 2
B = 79 679 68 79 9 79 78 9 89 8 89 89 9 8 9 9 9
E = 11 111 11 11 2 11 22 4 44 3 55 55 5 5 5 6 6
1 11 2 33 2 33 44 3 4 4 5 5
1 11 1 11 22 2 2 3 2 3
1 1 1 1 1
G = 99 555 88 99 0 33 88 2 66 7 99 66 3 8 1 0 8
9 77 0 55 6 77 55 1 7 0 9 7
9 33 5 55 33 0 5 9 6 5
9 4 7 5 3
In several permutations, G conflicts with other letters. I'll mark those with x's:

x x x xx
A = 3x 333 3x 4x 4 xx 44 6 66 6 77 77 7 7 8 8 8
H = 2x 222 2x 3x 3 xx 33 5 55 5 66 66 6 6 7 7 7
D = 6x 888 9x 2x 7 xx 99 3 77 9 22 33 4 9 2 3 4
J = 8x 444 7x 8x 8 xx 66 8 22 4 44 11 8 3 6 4 2
B = 7x 679 6x 7x 9 xx 78 9 89 8 89 89 9 8 9 9 9
E = 1x 111 1x 1x 2 xx 2x 4 xx 3 5x xx 5 x 5 6 x
x x1 2 xx x xx 44 3 x 4 x x
x 11 x 11 xx 2 2 x 2 3
x 1 x 1 1
G = 9x 555 8x 9x 0 xx 8x 2 xx 7 9x xx 3 x 1 0 x
x x7 0 xx x xx 55 1 x 0 x x
x 33 x 55 xx 0 5 x 6 5
x 4 x 5 3
Let me list the remaining permutations (removing the x's):

A = 3 333 3 4 4 44 66 66 6 777 77 777 77 88 888 88
H = 2 222 2 3 3 33 55 55 5 666 66 666 66 77 777 77
D = 6 888 9 2 7 99 33 77 9 222 33 444 99 22 333 44
J = 8 444 7 8 8 66 88 22 4 444 11 888 33 66 444 22
B = 7 679 6 7 9 78 99 89 8 889 89 999 88 99 999 99
E = 1 111 1 1 2 21 42 11 3 511 44 532 21 54 621 31
G = 9 555 8 9 0 87 20 33 7 955 55 310 54 10 065 53
H+I=10+E or H+I=9+E (if there is a borrow from E). So we can add I to this list:

A = 3 333 3 4 4 44 66 66 6 777 77 777 77 88 888 88
H = 2 222 2 3 3 33 55 55 5 666 66 666 66 77 777 77
D = 6 888 9 2 7 99 33 77 9 222 33 444 99 22 333 44
J = 8 444 7 8 8 66 88 22 4 444 11 888 33 66 444 22
B = 7 679 6 7 9 78 99 89 8 889 89 999 88 99 999 99
E = 1 111 1 1 2 21 42 11 3 511 44 532 21 54 621 31
G = 9 555 8 9 0 87 20 33 7 955 55 310 54 10 065 53
I = 99 7 8 5 5 5 6
Notice that only eight columns work. So we have reduced the permutations to eight. And we have looked at eight of the letters. The other two letters (C and F) can only have a couple of possible values each. Here is a list of all the possible values of AxBCD=HICJ:

678 768 943 923 934 829 913 914
x3 x3 x6 x7 x7 x7 x8 x8
---- ---- ---- ---- ---- ---- ---- ----
2034 2304 5658 6461 6538 5803 7304 7312
Six of these can be eliminated immediately (because C is not the same above and below), and one more (the right one) can be eliminated with a little checking. And that leaves only one permutation. And this is the solution:

7
----
934)7210
6538
----
672


This may looks complicated but analized it

Answer 3

Okay, I am in geometry, and I do not know if this is even close, but I'll give it a try. First,

(X-9) (X3-8) (2X-8) = 0

Okay, first I took the problem and I used the distributive property, (like I said before, I have never been presented with a question on this scale) and I wrote the whold dern thing out as follows, (may get confusing)

X5 - X2 -8 - 9x3 + 73 + 2x2 - 73 + 2x3 - x3 - 8 - 42

Then I took the exponents and I got them together, and I took the whole numbers with out exponents and got them together, to where it looked like this,

x5 - x2 -9x3 +2x2+2x3 -x3 ------------ -8 + -73 +-50 -131

So I then took these numbers and got them together, presenting them in the following text,

x5 + x2 -8x3 -131 = 0

I reduced and simplified so it looked like

x7 - 8x3 - 131 = 0

I subtracted 131 from each side, and I got my semi-final answer

x7 - 8x3 = 131 So I leave you with the following,

I hate math, I do what I can to get by, I try and try to do better but I cannot, and I hope that during the time that I was figuring out this problem and finding a completely wrong answer that someone with a larger brain can decipher this great mathematical problem, do it in their head, and give you the answer.

But the last few digits should help, if I did it right.

Answer 4

I did this last year. Sorry, I'm too tired to come up with the answer but I can tell you how to do it. You have to multiply x2 by x3 1st. Then, you muliply x2 by -8 and subract it from the first number. I can't go on, but I hope I helped you to remember what you probably learned in class. Good luck!

Answer 5

x2-9=0
or
x3-8=0
or
2x-8=0

x = 3, -3, 2, 4

Answer 6

(x^2 - 9)(x^3 - 8)(2x - 8) = 0

in order for the answer to be zero, either
(x^2 - 9) = 0
or
(x^3 - 8) = 0
or
(2x - 8) = 0
............................
x^2 - 9 = 0
x^2 = 9
x= +/-3.........first solution
-----------------------
x^3 - 8 = 0
x^3 = 8
x= 2..........second solution
.............................
2x - 8 = 0
2x = 8
x= 4



Therefore x = 2, +/-3, or 4

Answer 7

x=3, -3, 4, and 2

I put it into Microsoft Math