/sci/ - Science & Math » Thread #14331445

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Anonymous Thu 24 Mar 14:54:35 2022 No.14331445 View Reply Original Report

Quoted By:

what is 4x^2 - 3y^2 ?

Anonymous

Anonymous Fri 25 Mar 2022 04:48:21 No.14333869 Report

Quoted By: >>14333937

asdf

Anonymous

Anonymous Fri 25 Mar 2022 05:18:18 No.14333937 Report

Quoted By:

>>14333869
Just get to zero anon. Find like terms and cross them out until you get a new function. That new function will equal zero on (x,y) or (0,0)

Anonymous

Anonymous Fri 25 Mar 2022 05:19:41 No.14333942 Report

Quoted By: >>14333951

$a =\left( 2 + \sqrt{3} \right)^{1997}$

$b =\left( 2 - \sqrt{3} \right)^{1997}$

$x = \frac{a+b}{2}, \ \ y = \frac{a - b}{\sqrt{3}}$

$4x^2 - 3y^2 = 4\left(\frac{a+b}{2}\right)^2 - 3\left(\frac{a - b}{\sqrt{3}}\right)^2$

$4\frac{\left(a^2 + b^2 + 2ab\right)}{4} - 3\frac{\left(a^2 + b^2 - 2ab \right )}{3}$

$4x^2 - 3y^2 = 4ab = 4\left(\left( 2 + \sqrt{3} \right)^{1997}\left( 2 - \sqrt{3} \right)^{1997} \right ) =\\<br /><br />=\left(4 - 2\sqrt{3} +2\sqrt{3} - 3 \right ) = 1$

Anonymous

Anonymous Fri 25 Mar 2022 05:22:02 No.14333950 Report

Quoted By: >>14333964

correction on the last line

$4x^2 - 3y^2 = 4ab = 4\left(\left( 2 + \sqrt{3} \right)^{1997}\left( 2 - \sqrt{3} \right)^{1997} \right ) =4\left(4 - 2\sqrt{3} +2\sqrt{3} - 3 \right )^{1997} = 4$

Anonymous

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Anonymous Fri 25 Mar 2022 05:22:09 No.14333951 Report

Quoted By: >>14333964

>>14333942
I kneel

Anonymous

Anonymous Fri 25 Mar 2022 05:25:20 No.14333964 Report

Quoted By: >>14334443

>>14333951
the answer is 4, not 1, check >>14333950

/sci/ should have a LaTeX preview

Anonymous

Anonymous Fri 25 Mar 2022 09:21:26 No.14334443 Report

Quoted By:

>>14333964
correct

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