Question

The sum of two numbers is \(\color{green}{6}\) times their \(\color{blue}{Geometric Mean}\).

Show that the Numbers are in the Ratio -
\(\color{red}{(3+2\sqrt{2}) : (3 - 2 \sqrt{2})}\)

Answer 1

2\{sqrt
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Answer 2

Oh man where is the problem..

Answer 3

Oh, looks good now! :)

Answer 4

Oh sorry forgot to refresh the page,..

Answer 5

Let us come to the question now...

Answer 6

\[a+b = 6 \sqrt{ab}\]

Answer 7

\[\frac{a}{b} = \frac{6 \sqrt{a}}{\sqrt{b}} - 1\]

Answer 8

\[\frac{a}{b} = \frac{(3+2\sqrt{2})}{ (3 - 2 \sqrt{2})}\]

Answer 9

This, I think we have to prove..

Answer 10

I know

Answer 11

Yeah, I know too...

Answer 12

I'm pretty sure you can figure it out. Its just algebraic manipulation.

Answer 13

Let me try then..

Answer 14

I am not getting where to start from..

Should I divide both the sides by sqrt{ab} ??

Answer 15

I figured it out without doing any paperwork

Answer 16

That is why you are Hero I think..

Answer 17

\[\frac{a}{b} = 17 + 12 \sqrt{2}\]

Answer 18

?

Answer 19

this is some more simplification of what we have to prove..

Sorry not we, but me..

Answer 20

I think there is an easier way bro

Answer 21

Wait, I am thinking then...

Yeah there must be..

Answer 22

Just let me know when you give up

Answer 23

Sure why not..

Answer 24

Did you figure it out yet?

Answer 25

No, I am still trying it on my notebook..

Answer 26

Sorry, internet got disconnected..

Answer 27

Give me some hint, I am not getting where to start from..

Answer 28

Give me some hint if you can, I am not getting where to start from..

Answer 29

well this ain;t much of the problem..

you already know
a/b = 6sqrt(a/b) -1

just let sqrt(a/b) = x.. hope this helps..

Answer 30

I am hoping too..

Wait then, I will go forward now..

Answer 31

\[x = 3 \pm 2 \sqrt{2}\]

Answer 32

\[\frac{\sqrt{a}}{\sqrt{b}} = 3 \pm 2 \sqrt{2}\]

Answer 33

@hero Am I right??

Answer 34

\[\frac{\sqrt{a}}{\sqrt{b}} = 3 + 2 \sqrt{2}\]

\[\frac{\sqrt{a}}{\sqrt{b}} = 3 - 2 \sqrt{2}\]

Answer 35

If you let:\[a+b=k\]then it follows that:\[6\sqrt{ab}=k\Longrightarrow ab=\frac{k^2}{36}\]Therefore you want to solve this system:\[a+b=k\]\[ab=\frac{k^2}{36}\]which is the same as solving the quadratic:\[x^2-kx+\frac{k^2}{36}=0\] you can use the quadratic formula formula from here and divide the answers to get the desired ratio.

Answer 36

Wait I am finding here x as you are saying...

Answer 37

\[x = \frac{k \pm \sqrt{k^2 - \frac{k^2}{9}}}{2}\]

Answer 38

thats right :) so let a be the solution with the +, and b be the solution with the -.

Clean it up a little be too. You can factor the k^2 inside the radical, etc etc.

Answer 39

Yeah I am, Net is very slow..

Answer 40

\[x = \frac{3k \pm 2k \sqrt{2}}{2}\]

Answer 41

Wait there is 6 in the denominator...

Answer 42

nice, so if:\[a=\frac{3k+2k\sqrt{2}}{2}\]and\[b=\frac{3k-2k\sqrt{2}}{2}\] then:\[\frac{a}{b}=\]

Answer 43

I am not getting this that how you have taken a and b from here??

Answer 44

Ah, im sorry, i should have explained it better. So why is solving the system:\[a+b=k\]\[ab=\frac{k^2}{36}\]the same as solving the equation:\[x^2-kx+\frac{k^2}{36}=0\]?

Its because when you are solving the quadratic equation:\[x^2-bx+c=0\]by factoring, you ask yourself, "What are two numbers that add up to b, and multiply to c?" But this statement is exactly the system:\[m+n=b\]\[mn=c\] basically, solving the system of equations, or just the quadratic yield the same solutions.

That is why im saying the solutions to the quadratic are now a and b.

Answer 45

Oh so a and b are the roots of the equation...

Answer 46

exactly :)

Answer 47

I did not look that way..

Sorry I got it..

Thanks you for explaining me better in this regard...

Answer 48

*Thank..

And also thanks to @Hero

Sorry I could not follow you...