A Tale of Two Ratios #maths #algebra #education

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[mikehood]

I, too, like to find (1/x) when required to solve for x 🙃.

[HenkVanLeeuwen-io]

So 2a+1/a=2.
So a=0.5+0.5 sqrt 3 and b=1+sqrt 3 or a=0.5-0.5 sqrt 3 and b=1-sqrt 3.

[ethanbartiromo]

2a^2 - 2a + 1 = 0 so a = (1 +/- i)/2

So b = 1 +/- i

Verifying we get 1 + i + 2/(1 + i) = 1 + i + 2(1 - i)/2 = 2

And (1 + i)/2 + 1/(1 + i) = (1 + i)/2 + (1 - i)/2 = 1

(1 - i) + 2/(1 - i) = 1 - i + 2(1 + i)/2 = 2

And (1 - i)/2 + 1/(1 - i) = (1 - i)/2 + (1 + i)/2 = 1

So this works

[mircoceccarelli]

a, b =/= 0 => b = 2a

2a + 1/a = 2 => 2a^2 - 2a + 1 = 0

D = ( - 2 )^2 - 4( 2 )( 1 ) = 4 - 8 = - 4 < 0

No soluzioni reali

a = ( 1 +/- i )/2 => b = 1 +/- i

( a, b ) = { ( ( 1 + i )/2, 1 + i ),

( ( 1 - i )/2, 1 - i ) }

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