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 April 20th, 2014, 08:25 AM #1 Member     Joined: Sep 2013 From: New Delhi Posts: 38 Thanks: 0 Math Focus: Calculus solve it if u can calculate x and y: (2x^2)+(7y^2)+(2 multiplied by square root of 10 multiplied by x)+(2 multiplied by square root of 21 multiplied by y)+8=0
 April 20th, 2014, 11:04 AM #2 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 Use completing the square. You'll get an equation of the form $2(x+x_0)^2+7(y+y_0)^2=0$ for some $x_0,y_0$. Therefore, the solution is $x=-x_0$ and $y=-y_0$.
 April 20th, 2014, 11:51 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,484 Thanks: 2041 By the above method, I got (x, y) = (-sqrt(10)/2, -sqrt(21)/7).
 April 20th, 2014, 11:56 AM #4 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 That's correct. Edit: Ah, you are not the OP...
April 20th, 2014, 09:29 PM   #5
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Quote:
 Originally Posted by aditya ranjan calculate x and y: (2x^2)+(7y^2)+(2 multiplied by square root of 10 multiplied by x)+(2 multiplied by square root of 21 multiplied by y)+8=0
$\displaystyle y_1,y_2=\frac{1}{7} [-7y_0\mp i\sqrt{14}(x+x_0) ]$ where $\displaystyle i^2=-1$ and $\displaystyle x_0\in \mathbb C$.

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