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April 20th, 2014, 09:25 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, 12:04 PM  #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, 12:51 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,264 Thanks: 1959 
By the above method, I got (x, y) = (sqrt(10)/2, sqrt(21)/7).

April 20th, 2014, 12:56 PM  #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, 10:29 PM  #5 
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  $\displaystyle y_1,y_2=\frac{1}{7} [7y_0\mp i\sqrt{14}(x+x_0) ]$ where $\displaystyle i^2=1$ and $\displaystyle x_0\in \mathbb C$.
