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Unformatted text preview: terms.aspx College Algebra So, in this case that means that we only need to complete the square on the x terms. Recall how
this is done. We first take half the coefficient of the x and square it.
ç ÷ = ( 4 ) = 16
è2ø We then add this to both sides of the equation. x 2 + 8 x + 16 + y 2 = -7 + 16 = 9
Now, the first three terms will factor as a perfect square. ( x + 4) 2 + y2 = 9 Step 3 : This is now the standard form of the equation of a circle and so we can pick the center
and radius right off this. They are, center = ( -4, 0 ) radius = 9 = 3
[Return to Problems] (b) x 2 + y 2 - 3 x + 10 y - 1 = 0
In this part we’ll go through the process a little quicker. First get terms properly grouped and
placed. x 2 - 3x + y 2 + 10 y = 1
14243 14 3
complete the square complete the square Now, as noted above we’ll need to complete the square twice here, once for the x terms and once
for the y terms. Let’s first get the numbers that we’ll need to add to both sides.
2 2 æ 3ö 9
ç- ÷ =
è 2ø 2
æ 10 ö
ç ÷ = ( 5 ) = 25
è2ø Now, add these to both sides of the equation. 9
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
- Spring '12