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Let a and b with a>b>0 be real numbers satisfying a^2+b^2=4ab. Find a/b - b/a.

 Feb 19, 2022
 #1
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simplifying a/b-b/a gives a^2/ab - b^2/ab = \({a^2-b^2}\over ab\) which is the first equation divided by ab...except the subtraction sign... anyone else want to solve this? right now i'm trying to find a way to solve this..

 

 

also according to wolfram alpha a/b-b/a = \(\pm 2\sqrt{3}\)

 Feb 20, 2022
 #2
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hmm

\(a^2+2ab+b^2=6ab\\ (a+b)^2=6ab\\ a+b=\sqrt{6ab}\\ a^2-2ab+b^2=2ab\\ (a-b)^2=2ab\\ a-b=\sqrt{2ab}\)

So

 \(\frac{a^2-b^2}{ab}\\ =\frac{\sqrt{6ab}\cdot\sqrt{2ab}}{ab}\\ =\sqrt{12}\cdot\frac{ab}{ab}\\ =2\sqrt{3}\)

(notice that a and b are both positive so the answer must be positive)

Also, this question was asked quite a while ago:

https://web2.0calc.com/questions/algebra_16930

 Feb 20, 2022
 #3
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ahh, now i see/understand what you did (i was the first answer guest) 

 

nice job

Guest Feb 20, 2022

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