Completing the Square: How Does it Work?Date: 01/20/99 at 16:36:14 From: Dena Subject: Completing the Square I don't understand how to complete the square. Am I right that you are supposed to divide the second term by two and then square it? I always get messed up when there are fractions. Please help! Date: 01/23/99 at 20:04:23 From: Doctor Jeremiah Subject: Re: Completing the Square Hi Dena: Yes, you are right about the procedure. Let me try to show you why: When we complete the square the whole goal is to get the equation into a form so that it can be simplified into a square. Say we start with ax^2 + bx + c = 0. The idea is to translate that into (x + d)^2 = e. But (x + d)^2 = e when expanded is x^2 + 2dx + d^2 = e: ax^2 + bx + c = 0 ==> (x + d)^2 = e ==> x^2 + 2dx + d^2 = e The first step in this is to get rid of the factor on the x^2 term: ax^2 + bx + c = 0 ax^2/a + bx/a + c/a = 0/a x^2 + bx/a + c/a = 0 Now we substitute 2d for b/a because when we complete the square these terms must be equal: 2d = b/a d = b/(2a) d^2 = (b/(2a))^2 And if you look at the numeric term carefully, the c changes to d^2-e: c/a = d^2 - e e = d^2 - (c/a) <== d^2 = (b/(2a))^2 e = (b/(2a))^2 - (c/a) Now if we put it all together: ax^2 + bx + c = 0 ==> x^2 + bx/a + (b/(2a))^2 = (b/(2a))^2 - (c/a) And if you look at this you'll see that you were exactly right in saying that the third term is the second term divided by two and then squared. The important thing to notice is that the value on the right of the equals sign changes. For example, say the equation we have is: x^2 + 6x - 16 = 0 Then if we move the 16 to the right side: x^2 + 6x = 16 Now to find the real third term. (6/2)^2 = 9. But remember you need to add it to both sides: x^2 + 6x + 9 = 16 + 9 x^2 + 6x + 9 = 25 (x+3)^2 = 25 If you need me to explain better why it works, just write back. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/