Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

4 Answers

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

share|improve this answer
add comment

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

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.

share|improve this answer
why don't you leave these questions to us kids? :P –  Soham Chowdhury Jul 27 '13 at 4:45
add comment

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

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

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

share|improve this answer
Could you please explain your first step? Did you just choose to divide both sides by $2ab$? –  gekkostate 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
add comment

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

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.