# Algebra help please

1. I have two equations that I just can't figure out. Can someone look and see if they can tell me how to figure them?

16x-8=

x^2x - 6 = 0 then x is ?

I have to be able to solve these for the admission exam on Tuesday Thanks
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3. 16x=8
x=8/16
x=1/2

i don't know what the second equation means (is that saying x the the something power times 2x minus 6???

good luck!
4. Ty the second reads x(secondpower)X-6=0
5. Ok, then you're going to have to factor:

x2-x-6=0
(x-3)(x+2)=0
x=3 or x=-2

hope that helps!

kim
6. at the risk of looking really dumb (x-3)(x+2)=0 where did this come from?
7. Originally posted by Lisa1970
Ty the second reads x(secondpower)X-6=0
Hmmm...Wouldn't zero work?

0^0-6=0

Zero to the power of anything will always be zero. At least that's what I think would be the case. It's been awhile since I did this stuff...
8. Okay, I don't think I understood the problem correctly. Ignore my answer because I am obviously more confused than anyone else...
9. 2ner has the right answer according to my answer sheet, but I am still unclear on how she did it
10. Is the equation x2-x-6=0? (for some reason that's what I assumed it was since I didn't see a sign in between the exponent and the 2nd x.

If so, you hafta factor....

(x-3)(x+2) is equal to x2-x-6 you can figure it out by foiling the (x-3)(x+2).

after you foil you get x2+2x-3x-6 which is the same as x2-x-6

then you have two terms that equal zero
you have x-3=0 AND x+2=0
then you have to isolate the x's by adding three to both sides of the first term you get x=3 and by subtracting the 2 from both sides of the second term you get x=-2

If the equation is x2 + x - 6 = 0 then it factors out to be (x+3)(x-2) and then the answers would b x=-3 or x=2

does that make sense?

kim
Last edit by 2ner on Aug 7, '03
11. You have to use logs for the second problem.

x^2x -6=0
x^2x=6
logx^2x=log6
2x (logx)=log6

I don't have my calculator with me right now but then plug it in and then you can solve for x. I hope that helps (and I hope that's right )

Keely
12. Okay, never mind. I thought the problem was x to the power of 2x minus 6. Whoops.

Keely
13. the exact equation is x small 2 (above x) x - 6 = 0, then x is

X2 x - 6 = 0, then x is
14. So 2nr has the correct answer? If so, then there is a misprint in this problem (actually in both...but I'll get there). If the answer to the second problem is x=3 or x=(-2) then the problem should read the way 2nr interpreted it: x(squared) - x - 6 = 0. It seems like you are missing a minus sign in the original problem.

As for the first problem, the answer is only 1/2 if the equation is equal to 0. (i.e. 16x - 8 = 0). In the way it is written above, it is actually unsolvable.

PJ

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