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# In Response to Soner Karaca

This is in response to Soner Karaca, who asked me the following questions. Everyone is free to provide feedback.

Problem:

1. Given $$a^2+b^2+ab-\sqrt 3b+1=0$$, find $$b-a$$.
2. Find $$7^{77} \text{ mod } 1000$$.

Note by Chew-Seong Cheong
2 weeks ago

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Solution to Problem 1

\begin{align} a^2+b^2+ab-\sqrt 3b+1 & =0 \\ b^2 + (a-\sqrt 3)b + a^2+ 1 & = 0 \\ \implies b & = \frac {\sqrt 3-a \pm \sqrt{a^2-2\sqrt 3a + 3 - 4a^2-4}}2 \\ & = \frac {\sqrt 3-a \pm \sqrt{-3a^2-2\sqrt 3a -1}}2 \\ & = \frac {\sqrt 3-a \pm (\sqrt3 a+1)i}2 \\ \implies b - a & = \frac {\sqrt 3-3a \pm (\sqrt3 a+1)i}2 \end{align} · 1 week, 6 days ago

sir, what is this note about? I mean is this note written to tell the solution for for some other purpose.. Although I have understood the solution you wrote of the above problem. · 2 weeks ago

I was replying to @Soner Karaca who asked me this problem through Messenger. I need LaTex to reply him. · 1 week, 6 days ago

ok sir. · 1 week, 6 days ago

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