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:

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

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Solution to Problem 1

$\begin{aligned} a^2+b^2+ab-\sqrt 3b+1 & =0 \\ b^2 + (a-\sqrt 3)b + a^2+ 1 & = 0 \\ \implies b & = \frac {\sqrt 3-a \pm \sqrt{a^2-2\sqrt 3a + 3 - 4a^2-4}}2 \\ & = \frac {\sqrt 3-a \pm \sqrt{-3a^2-2\sqrt 3a -1}}2 \\ & = \frac {\sqrt 3-a \pm (\sqrt3 a+1)i}2 \\ \implies b - a & = \frac {\sqrt 3-3a \pm (\sqrt3 a+1)i}2 \end{aligned}$

Chew-Seong Cheong - 4 years, 6 months 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.

Rakshit Joshi - 4 years, 6 months ago

I was replying to @Soner Karaca who asked me this problem through Messenger. I need LaTex to reply him.

Chew-Seong Cheong - 4 years, 6 months ago

ok sir.

Rakshit Joshi - 4 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$