Completing the Square to Solve Quadratic EquationsDate: 12/21/98 at 12:56:11 From: Jim Subject: Quadratic Equations My textbook is attempting to tell me (I am homeschooled) the two ways to solve quadratic equations. We start with: x^2 + x - 2 = 0 I already know how to factor it out to: (x + 2)(x - 1) = 0 x = -2 OR x = 1 What I don't know is how they turned it into the form: (x - 2)^2 = 3 I have an idea how they got there, something like b = sqrt(a^2 +c) all divided by 2b,) but looking back I can't find it anywhere. I don't even know where to begin to solve this. Date: 12/21/98 at 16:37:19 From: Doctor Rob Subject: Re: Quadratic Equations Thanks for writing to Ask Dr. Math! The procedure is called, "Completing the Square." It begins with knowing that (r+s)^2 = r^2 + 2*r*s + s^2. Starting with x^2 + x - 2 = 0 we want to build a square on the left-hand side by using the x^2 and x terms. To do this we start by adding 2 to both sides to isolate these terms on the left: x^2 + x = 2 Then we multiply by 4 times the coefficient of x^2, which is 4*1 = 4 in this case: 4*x^2 + 4*x = 8 Now set the first two terms of the square of r + s equal to the two terms on the left-hand side: r^2 = 4*x^2 and 2*r*s = 4*x. Then r = 2*x and s = 1 works. (It is also true that r = -2*x and s = -1 works. You can use either pair.) Now add s^2 = 1^2 = 1 to both sides: 4*x^2 + 4*x + 1 = 8 + 1 = 9 Now the left side is a perfect square, since it equals r^2 + 2*r*s + s^2, which equals (r+s)^2, or (2*x+1)^2. This is so because we carefully arranged for it to be so. Also the right-hand side 9 is a square, namely the square of 3: (2*x + 1)^2 = 3^2. This is the corrected equation that you mentioned above as the one "they turned it into." Now take the square root of both sides, remembering that either of them could be negative: 2*x + 1 = 3 or 2*x + 1 = -3, 2*x = 2 or 2*x = -4, x = 1 or x = -2. That is how to use "Completing the Square" to solve this quadratic equation. If you start with the general quadratic equation, a*x^2 + b*x + c = 0 where a, b, and c are any expressions with a not zero, the same steps are followed, and you get the following equations in turn: a*x^2 + b*x = -c, 4*a^2*x^2 + 4*a*b*x = -4*a*c, r^2 = 4*a^2*x^2, 2*r*s = 4*a*b*x, r = 2*a*x, s = b, s^2 = b^2, 4*a^2*x^2 + 4*a*b*x + b^2 = b^2 - 4*a*c, (2*a*x + b)^2 = b^2 - 4*a*c, 2*a*x + b = sqrt(b^2-4*a*c) or -sqrt(b^2-4*a*c), 2*a*x = -b + sqrt(b^2-4*a*c) or -b - sqrt(b^2-4*a*c), x = (-b+sqrt[b^2-4*a*c])/(2*a) or x = (-b-sqrt[b^2-4*a*c])/(2*a) The last two lines are called the "Quadratic Formula." It is useful to memorize this. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/