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April 20th, 2014, 08:25 AM   #1
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Math Focus: Calculus
Exclamation 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
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April 20th, 2014, 11:04 AM   #2
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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 $y=-y_0$.
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April 20th, 2014, 11:51 AM   #3
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By the above method, I got (x, y) = (-sqrt(10)/2, -sqrt(21)/7).
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April 20th, 2014, 11:56 AM   #4
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That's correct.

Edit: Ah, you are not the OP...
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April 20th, 2014, 09:29 PM   #5
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Quote:
Originally Posted by aditya ranjan View Post
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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