Please;

how do I get $\displaystyle \sqrt{x+1}=x\sqrt{}1/x^2-1$

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- Jan 12th 2013, 06:01 AMmarijakopljarRoot-simple,but...
**Please;**

how do I get $\displaystyle \sqrt{x+1}=x\sqrt{}1/x^2-1$ - Jan 12th 2013, 07:28 AMkalyanramRe: Root-simple,but...
- Jan 12th 2013, 05:10 PMProve ItRe: Root-simple,but...
How do you get this equation? How should we know?

Perhaps you're really asking how do you SOLVE this equation? - Jan 12th 2013, 06:08 PMHallsofIvyRe: Root-simple,but...
What you wrote is the same as $\displaystyle \sqrt{x+1}= x/x^2- 1= (1/x)- 1$. What you perhaps intended was $\displaystyle \sqrt{x+ 1}= x\sqrt{\frac{1}{x^2- 1}}= \frac{x}{\sqrt{x^2- 1}}$

But "how do I get" an equation usually means "how do I show that these are the same for all x". And what you have**isn't**. For example if x=3, the left side (of both of those) is $\displaystyle \sqrt{4}= 2$ while the right side of the last is $\displaystyle \frac{3}{\sqrt{9- 1}}= \frac{3\sqrt{2}}{4}$, NOT 2. For the first, it is 1/3- 1= -2/3.

If you mean "how do I solve $\displaystyle \sqrt{x+ 1}= x\sqrt{\frac{1}{x^2- 1}$?} the simplest thing to do is get rid of the square roots by squaring both sides: $\displaystyle x+ 1= \frac{x^2}{x^2- 1}$ which, mutiplying both sides by $\displaystyle x^2- 1$ gives $\displaystyle (x+ 1)(x^2- 1)= x^3+ x^2- x- 1= x^2$ so you have to solve the cubic equation $\displaystyle x^3- x- 1= 0$. However, that has no rational number roots so I doubt if that is what you mean. - Jan 13th 2013, 03:08 AMibduttRe: Root-simple,but...
There is some doubt in the explanation. What do you want to do??