Major error in Barron's Math II 8th edition:

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1

<p>Page 123, part B of example #1:</p>

<p>(4x^2) + (4y^2) - 12x - 20y - 2 = 0</p>

<p>You're supposed to complete the square to find the equation of the circle and here's how they do it:</p>

<p>(x^2 - 3x + 9/4) + (y^2 - 5y + 25/4) - 2/4 = 9/4 + 25/4</p>

<p>(<em>note the 2/4</em>)</p>

<p>and get: </p>

<p>( x - 3/2)^2 + ( y - 5/2)^2 = 9</p>

<p>My question is: shouldn't the costant '2' of the original equation be carried out the other side of the equal sign which would change the answer completely? And why is it 2/4 instead of just 2? </p>

<p>...or is this just another of Barron's typos??</p>

2

<p>No, I believe they have it right.</p>

<p>Divide both sides of the original by 4 to get: x^2 + y^2 - 3x - 4y - 2/4 = 0 (the reason they left it in the form of 2/4 is because, in completing the square, you have to add terms which have a denominator of 4…so 2/4 makes the whole equation look a little nicer).</p>

<p>Complete the square by adding 9/4 and 25/4 to both sides: (x^2 - 3x + 9/4) + (y^2 - 5y + 25/4) - 2/4 = 9/4 + 25/4</p>

<p>Simplify and take 2/4 to the right: (x-1.5)^2 + (y-2.5)^2 = 9/4 + 25/4 + 2/4 = 36/4 = 9</p>

3

<p>4x^2+4y^2-12x-20y-2=0</p>

<p>4x^2+4y^2-12x-20y-2+9+25=9+25 (just add 9 and 25)</p>

<p>x^2+y^2-3x-20y-2/4+9/4+25/4=9/4+25/4 (divide everything by four)</p>

<p>x^2-3x+9/4+y^2-20y+25/4=9/4+25/4+2/4 (rearrange them and add 2/4 to both sides)</p>

<p>(x-2/3)^2+(y-5/2)^2=9 (factor out the left and add up the right)</p>

4

<p>Even if you took it to the other side first, you’d get the same result. Take it to other side first, then divide by 4, then complete the square. Same thing.</p>

5

<p>I concur. They’ve got it correct.</p>

6

<p>THanks for all the help!</p>