One of many situations in which you might need to complete the square is the following:<\/p>\n
x^2 – 6x =17<\/p>\n
It’s a quadratic; naturally, you want to subtract 17 from both sides and hopefully factor<\/a>.<\/p>\n x^2 – 6x -17 = 0<\/p>\n However, you already know that x2<\/sup> – 6x -17 doesn’t factor. What might you do instead?<\/p>\n Completing the square is one option. To do so, leave 17 on the right side:<\/p>\n x^2 -6x =17<\/p>\n Step 1:<\/p>\n Focus on the coefficient of the x term: in this case, the -6. Specifically, take half of it, then square that.<\/p>\n -6\u00f72 = -3→(-3)^2=9<\/p>\n Step 2:<\/p>\n Add the result from Step 1 to both sides of the equation:<\/p>\n x^2 – 6x + 9 = 17 + 9<\/p>\n Step 3:<\/p>\n Notice that now, the left side does indeed factor<\/a>. In this case, it factors to (x-3)(x-3), aka (x-3)2<\/sup>. We rewrite the equation correspondingly:<\/p>\n (x-3)^2=26<\/p>\n Step 4:<\/p>\n With the left side in square form, we can square root both sides, to get<\/p>\n x-3=±√(26)<\/p>\n We add three to both sides, yielding<\/p>\n x=3+√(26) or x=3-√(26)<\/p>\n This is a light introduction to a topic that can be more complicated. However, it shows the usefulness and elegance of completing the square. More uses and complications of it will be visited in future posts:)<\/p>\n