THE SOLUTION OF

[X + (X^2-1)^1/2 / X - (X^2-1)^1/2] + [ X - (X^2-1)^1/2 / X + (X^2-1)^1/2 ] = 14

a) +8

b) -6

c) + or- 2

d) + or - 4

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- Jun 13th 2013, 12:11 PMsscgeekImportant question plzzzzzzz help
THE SOLUTION OF

[X + (X^2-1)^1/2 / X - (X^2-1)^1/2] + [ X - (X^2-1)^1/2 / X + (X^2-1)^1/2 ] = 14

a) +8

b) -6

c) + or- 2

d) + or - 4 - Jun 13th 2013, 12:16 PMMarkFLRe: Important question plzzzzzzz help
I have a few suggestions for you to get prompt help here:

- Use unique and descriptive topic titles.
- Use proper bracketing symbols or better yet use $\displaystyle \LaTeX$ to make your questions unambiguous.
- Show you work so the helpers here know exactly where you are stuck.

- Jun 13th 2013, 12:41 PMebainesRe: Important question plzzzzzzz help
What you wrote is this:

$\displaystyle \left [ x+ \frac {\sqrt{x^2-1}} x - \sqrt {x^2-1} \right ] + \left [ x - \frac {\sqrt{x^2-1}} x + \sqrt {x^2-1} \right ] = 14 $

Note that many of the terms cancel out, leaving just 2x = 14. So it seems that for the equation you wrote none of the answers are correct. - Jun 13th 2013, 03:13 PMSorobanRe: Important question plzzzzzzz help
Hello, sscgeek!

Quote:

$\displaystyle \text{Solve: }\: \frac{x + \sqrt{x^2-1}}{x - \sqrt{x^2-1}} + \frac{x - \sqrt{x^2-1}}{x + \sqrt{x^2-1}}\;=\;14$

. . $\displaystyle (a)\;+8 \qquad (b)\;-6 \qquad (c)\;\pm 2 \qquad (d)\; \pm4$

We have: .$\displaystyle \frac{(x+\sqrt{x^2-1})^2 + (x-\sqrt{x^2-1})^2} {(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})} \;=\; 14 $

. . $\displaystyle \frac{(x^2 + 2x\sqrt{x^2-1} + x^2-1) + (x^2 - 2x\sqrt{x^2-1} + x^2-1)}{x^2 - (x^2-1)} \;=\;14 $

. . $\displaystyle \frac{4x^2 - 2}{1} \:=\:14 \quad\Rightarrow\quad 4x^2 \:=\:16 \quad\Rightarrow\quad x^2 \:=\:16$

. . $\displaystyle x \:=\:\pm4$ .Answer (d)

- Jun 13th 2013, 03:54 PMHallsofIvyRe: Important question plzzzzzzz help
Interesting. Soroban's interpretation of what sscgeek wrote is quite different from ebaines' interpretation.

- Jun 14th 2013, 07:20 AMjpritch422Re: Important question plzzzzzzz help
I would say that Soroban's interpretation is correct (paretheses needed in the equation), though I can see how ebaines could interpret the equation the way he did without the parentheses.

- Jun 14th 2013, 09:11 AMtopsquarkRe: Important question plzzzzzzz help