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For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$

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up vote 8 down vote accepted

We have $a^2+b^2=6ab$.

To both sides, add $2ab$ to obtain, $(a+b)^2 = 8ab$.

Similarly, subtract $2ab$ to obtain, $(a-b)^2=4ab$.

Thus, $\left(\dfrac{a+b}{a-b}\right)^2 = 2$.

So ultimately $\frac{a+b}{a-b}= \pm\sqrt2$.

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$$(a-b)^2=4 a b = (a+b)^2-(a-b)^2$$

which means that

$$1 = \left ( \frac{a+b}{a-b}\right)^2 - 1$$


$$\frac{a+b}{a-b} =\pm \sqrt{2}$$

depending whether $a > b$ or not.

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why don't you leave these questions to us kids? :P – Soham Chowdhury Jul 27 '13 at 4:45

We have $$\frac{a^2+b^2}{2ab}=\frac31$$

Applying componendo and dividendo, $$\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{3+1}{3-1}$$

$$\implies \left(\frac{a+b}{a-b}\right)^2=2$$

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Could you please explain your first step? Did you just choose to divide both sides by $2ab$? – Jeel Shah Jul 27 '13 at 4:46
Yes. ${}{}{}{}$ – The Chaz 2.0 Jul 27 '13 at 5:48
@gekkostate, yes, then applied componendo and dividendo – lab bhattacharjee Jul 27 '13 at 6:03

Note that you would have $ \ (a + b)^2 = a^2 + 2ab + b^2 = 8ab \ , $ and something similar for $ \ ( a - b )^2 \ $ ...

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