Completing the Square of a Polynomial
Date: 5/22/96 at 19:4:55 From: Anonymous Subject: How do you complete the square? I have no idea what completing the square is. I know how to factor, but I don't understand this concept. Please help.
Date: 5/28/96 at 19:31:20 From: Doctor Pete Subject: Re: How do you complete the square? Completing the square is a lot like factoring. Say you have a polynomial like x^2-2x+4 . If you look at the first two terms, they are the same as x^2-2*x+1 = (x-1)^2. So if we write the 4 as 1+3, we get (x-1)^2+3. We've just completed the square of this polynomial. Now, let's try x^2-5x+1 . The 5 is a bit weird, since (x-2)^2 = x^2-4x+4, but (x-3)^2 = x^2-6x+9. So for what value of m will (x-m)^2 = x^2-5x+[something]? We multiply it out: (x-m)^2 = x^2-2mx+m^2. So 2m=5, or m=5/2. But this means that m^2 = 25/4, so if we are to complete the square, we need to do this: x^2-5x+(25/4-25/4)+1 = (x^2-5x+25/4)+(-25/4+1) = (x-5/2)^2-21/4. Now we have completed the square of this polynomial. The point of completing the square is to express some polynomial as a square plus some number. Now, try this: 3x^2+x+7 . Here we have an extra 3, so what do we do now? Well, we can divide by 3, complete the square, and then multiply it back to get 3((x+1/6)^2-85/36) . So you can't get rid of the 3. But say we have 3x^2+x+7=0, and we want to know what x will solve this equation. Since we have completed the square, this equation is the same as 3((x+1/6)^2-85/36)=0. Then we can divide out the 3 and move the 85/36 to the other side to get (x+1/6)^2 = 85/108 and now we can take the square root and subtract 1/6 to get x = sqrt(85/108) - 1/6 = (sqrt(85)/6sqrt(3)) - 1/6 = sqrt(255)/18 - 3/18 = (sqrt(255) - 3)/18. So we have a solution. (There is another, x = (-sqrt(85)-3)/6. Why?) So the cool thing about completing the square is that it allows you to solve equations like the one we just did here. In fact, if you haven't seen it already, it is the method by which one usually derives the quadratic formula; that is, if ax^2 + bx + c = 0, then -b +/- sqrt(b^2-4ac) x = ---------------------- . 2a -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum