Question

X^1-X^-1+x=2
A.0
B.1/2
C.11/2
D.1/2

Answer 1

can u draw your equation above, please

Answer 2

idk how to do this.. i need help so bad.. =(

Answer 3

In the reply window, there is a blue button for equation and one for drawing. Its on the left bottom box where you type your post.

Answer 4

\[x^1-x^{-1}+x=1\\
2x-x^{-1}=1\\
2x-\frac{1}{x}-1=0\\
x\ne 0\\
\mbox{both sides}\times x\\
2x^2-1-x=0\\
x^2-\frac{x}{2}-\frac{1}{2}=0
\]
Now you use the formula to solve for the second degree equation.

Answer 5

But I just realized. Its easier to just test out the values in the answers:
x=0, \(0^1-0^{-1}+0\ne1\) so its not A.

Answer 6

ahhh okays thanks

Answer 7

\[\large 2x^2 -x-1\]\[\large x^2-x-2\]\[\large (x-2)(x+1)\]\[\large (x-\frac{2}{2})(x+\frac12)\]\[\large (x-1)(2x+1)=0\] Hmmm.

Answer 8

First off, you have 2 of the same answers.Are you sure these are the right answers?

Answer 9

x=1/2, \(\displaystyle\left(\frac{1}{2}\right)^1-\left(\frac{1}{2}\right)^{-1}+\frac{1}{2}\)
\[\frac{1}{2}-2+\frac{1}{2}=1-2\ne2\] so its not B either.

Answer 10

Yeah, those repeated options must be wrong.

Answer 11

11/2 doesn't equal 2 either.

Answer 12

The answers would have been 1 and -1/2.

Answer 13

Yes. Those are the answers for the second degree equation too. Just like you factored.

Answer 14

Mmhmm.

Answer 15

I think the kid went away to school. He looked like he was doing his homework in a rush.

Answer 16

lol -_-

Answer 17

I think I made a mistake here
\[x^1-x^{-1}+x=1\\
2x-x^{-1}=1\\
2x-\frac{1}{x}-1=0\\
x\ne 0\\
\mbox{both sides}\times x\\
2x^2-1-x=0\\
x^2-\frac{x}{2}-\frac{1}{2}=0\]

Answer 18

completing the square?

Answer 19

ok,you ended up with the same equation though xD

Answer 20

It should be:
\[x^1-x^{-1}+x=2\\
2x-x^{-1}=2\\
2x-\frac{1}{x}-2=0\\
x\ne 0\\
\mbox{both sides}\times x\\
2x^2-1-2x=0\\
x^2-x-\frac{1}{2}=0\]

Answer 21

I think its still wrong. Should have used pen and paper.

Answer 22

Oh ok, it looks like it works now.

Answer 23

\[\large 2x -\frac1x -2=0\]\[\large x(2x-\frac1x-2=0)\]\[\large 2x^2 -1-2x =0\]\[\large 2x^2 - 2x - 1 = 0\]
0 does not equal 1
when x = 1/2 , -3/ 2 does not equal 0
when x= 11/2 , 97/2 does not = 0.

Answer 24

Yes, there is no right answer among those provided. Your alternatives are right: either 1 or -1/2.

Answer 25

Mmhmm.