Completing the Square in Vertex FormDate: 01/13/98 at 19:29:29 From: Carolyn Subject: Advanced Algebra (Algebra II) Dear Dr. Math, I am a sophmore taking Advanced Algebra. I am having trouble with these kinds of problems like this: y = 10x squared + 10x + 1 We are supposed to put it into vertex form (y-k = a(x-h)squared. I sort of get the process but I always make a different mistake, and even more important, I need to know why every step is done. I would appreciate it if your answer was in laymen's terms and work is shown. Thank you for your time. It is greatly appreciated. Sincerely, Carolyn Date: 01/26/98 at 11:02:57 From: Doctor Joe Subject: Re: Advanced Algebra (Algebra II) Dear Carolyn, I used to get the wrong answer for this type of problem too, but here's how I got over it: Let a, b, c be constants. Suppose we are looking at the expression y = ax^2 + bx + c. Step 1: Isolate the c from the ax^2 + bx term y = {ax^2 + bx} + c Step 2: Next, pluck out the a: y = a{x^2 + b/a} + c Step 3: Add the term a(b/2a)^2 to both sides. y + a(b/2a)^2 = a(x^2 + b/a + (b/2a)^2) + c Step 4: Complete the square. y + b^2/4a = a(x+b/2a)^2 + c Step 5: Bring the constant over to the y-side. y - (c - b^2/4a) = a(x + b/2a)^2 This is the required form. Now, by putting a = 10, b = 10, c = 1, you should be able to work out the right answr. The reason why you need this form is that it is extremely useful for solving the quadratic equation: ax^2 + bx + c = 0 Here we set y = 0 in the above "vertex" form (commonly known as the completing the square), -(c- b^2/4a) = a(x + b/2a)^2 Now, we have (x + b/2a)^2 = (b^2 - 4ac)/4a^2 So, we have x + b/2a = + sqrt{(b^2 - 4ac)/4a^2} or - sqrt{(b^2 - 4ac)/4a^2}. It follows that x = {-b + sqrt(b^2 - 4ac)}/2a or {-b - sqrt(b^2 - 4ac)}/2a. This is the famous quadratic formula for solving quadratic equations. (QED) P.S. Ever wonder why the form is called the "vertex" form? -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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