From 1 + ( (x-a˅1)/b )^2 < 1 + ( (x-a˅2)/b )^2

[subtle lantern]

In a formula from 1 + ( (x-a˅1)/b )^2 < 1 + ( (x-a˅2)/b )^2 I don't understand how to go further. The next step is 2x(a˅2 - a˅1) < a˅2^2 - a˅1^2. Then there's a note translated to As a˅2 > a˅1 then x < (a˅1 + a˅2) / 2. If you would kindly explain how to go across steps.

[cinder night]

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[latent minnow]

@subtle lantern Subtract 1 , multiply by b^2 (it is important that b^2 is positive) , expand the two squares , cancel the x^2 , and move the x to the left and the constant to the right , factor 2x and you should get to step 2.

From step 2 to step 3 divide everything to isolate the x, (this is why it is important to write $a_2 > a_1$ ), and on the right side you factor the difference of squares and you can cancel out the common term

[dense pumice]

AkselWellington

[subtle lantern]

About step 2 to step 3: I either misunderstood your explanation or forgot my understanding of it. AFAIK what you mean is 2x(a˅2 - a˅1) / 2(a˅2 - a˅1) < ( (a˅2 - a˅1)(a˅2 + a˅1) ) / 2(a˅2 - a˅1) then x < (a˅2 + a˅1) / 2(a˅2 - a˅1) but from that how do you get to step 3?

[latent minnow]

You should be able to can cancel out $a_2 - a_1$
Because you factor $a_2^2 - a_1^2 = (a_2 + a_1)(a_2 - a_1)$

[dense pumice]

AkselWellington