Prove an Equation involving Square Roots

**BabyMilo** · June 10th 2011, 10:56 AM

Proof $ \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1} $

fankyou.

**Quacky** · June 10th 2011, 11:08 AM

Originally Posted by BabyMilo

Proof $ \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^-1} $

fankyou.

I believe that there is a horrible typo in the question: it should be $\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } =\sqrt{a^2-1}$

I don't know if there is a shorthand, but I would consider writing the LHS as:
$\frac{1}{2}\sqrt{\frac{(a+1)(a-1)^2}{a-1}} +\frac{1}{2}\sqrt{\frac{(a-1)(a+1)^2}{a+1}}$

Take it from here.

Edit: I see you've corrected the typo in the OP

**BabyMilo** · June 10th 2011, 11:23 AM

I was trying to correct it, but my internet is so slow.
Thank you for your solution

**Soroban** · June 10th 2011, 12:05 PM

Hello, BabyMilo!

Another approach . . .

$\text{Prove: }\;\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } \;=\; \sqrt{a^2-1}$

We have: . $\frac{a-1}{2}\cdot\frac{(a+1)^{\frac{1}{2}}}{(a-1)^{\frac{1}{2}}} + \frac{a+1}{2}\cdot\frac{(a-1)^{\frac{1}{2}}}{(a+1)^{\frac{1}{2}}}$

. . . . . . . $=\;\frac{(a-1)^{\frac{1}{2}}(a+1)^{\frac{1}{2}}}{2} + \frac{(a+1)^{\frac{1}{2}}(a-1)^{\frac{1}{2}}}{2}$

. . . . . . . $=\;\frac{[(a-1)(a+1)]^{\frac{1}{2}}}{2} + \frac{[(a+1)(a-1)]^{\frac{1}{2}}}{2}$

. . . . . . . $=\;\frac{(a^2-1)^{\frac{1}{2}}}{2} + \frac{(a^2-1)^{\frac{1}{2}}}{2}$

. . . . . . . $=\;(a^2-1)^{\frac{1}{2}}$

. . . . . . . $=\;\sqrt{a^2-1}$

**gsmani9** · June 11th 2011, 12:04 PM

Originally Posted by BabyMilo

Proof $ \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1} $

fankyou.

L.H.S:-
$\frac{a-1}{2} \sqrt{\frac{a+1}{a-1}} + \frac{a+1}{2}\sqrt{\frac{a-1}{a+1}}$

$= \sqrt{\frac{(a+1)(a-1)(a-1)}{(a-1)4} } + \sqrt{\frac{(a-1)(a+1)(a+1)}{(a+1)4}}$

$= \sqrt{\frac{a^2-1}{4}} + \sqrt{\frac{a^2-1}{4}} = 2 \times \frac{1}{ 2}\sqrt{a^2-1}$

$= \sqrt{a^2 - 1}$

L.H.S. = R.H.S

I hope you have understood. All the best.

Thread: Prove an Equation involving Square Roots

LinkBack

Thread Tools

Display

Prove an Equation involving Square Roots

Equation involving square roots.

Similar Math Help Forum Discussions

A few questions involving square roots and inequalities

Inequality involving square roots

How do you calculate a Riemann sum involving square roots?

derivatives involving square roots

Evaluate involving arcsech and square roots

Search Tags